{"id":546136,"date":"2021-10-06T05:43:27","date_gmt":"2021-10-06T05:43:27","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546136"},"modified":"2021-10-07T05:21:31","modified_gmt":"2021-10-07T05:21:31","slug":"rd-sharma-solutions-for-class-8-maths-chapter-7-factorization","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-8-maths-chapter-7-factorization\/","title":{"rendered":"RD Sharma Solutions for Class 8 Maths Chapter 7\u2013Factorization"},"content":{"rendered":"\n

Class 8: Maths Chapter 7 solutions. Complete Class 8 Maths Chapter 7 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 7\u2013Factorization<\/h2>\n\n\n\n

RD Sharma 8th Maths Chapter 7, Class 8 Maths Chapter 7 solutions<\/p>\n\n\n\n

EXERCISE 7.1 PAGE NO: 7.3<\/h4>\n\n\n\n

Find the greatest common factor (GCF\/HCF) of the following polynomials: (1-14)<\/strong><\/p>\n\n\n\n

1. 2x2<\/sup> and 12x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 2 and 12<\/p>\n\n\n\n

The greatest common factor of 2 and 12 is 2<\/p>\n\n\n\n

The common literals appearing in given monomial is x<\/p>\n\n\n\n

The smallest power of x in two monomials is 2<\/p>\n\n\n\n

The monomial of common literals with smallest power is x2<\/sup><\/p>\n\n\n\n

\u2234 The greatest common factor = 2x2<\/sup><\/p>\n\n\n\n

2. 6x3<\/sup>y and 18x2<\/sup>y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 6 and18<\/p>\n\n\n\n

The greatest common factor of 6 and 18 is 6<\/p>\n\n\n\n

Common literals appearing in given numerical are x and y<\/p>\n\n\n\n

Smallest power of x in three monomial is 2<\/p>\n\n\n\n

Smallest power of y in three monomial is 1<\/p>\n\n\n\n

Monomial of common literals with smallest power is x2<\/sup>y<\/p>\n\n\n\n

\u2234 The greatest common factor = 6x2<\/sup>y<\/p>\n\n\n\n

3. 7x, 21x2<\/sup> and 14xy2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 7, 21 and 14<\/p>\n\n\n\n

Greatest common factor of 7, 21 and 14 is 7<\/p>\n\n\n\n

Common literals appearing in given numerical are x and y<\/p>\n\n\n\n

Smallest power of x in three monomials is 1<\/p>\n\n\n\n

Smallest power of y in three monomials is 0<\/p>\n\n\n\n

Monomials of common literals with smallest power is x<\/p>\n\n\n\n

\u2234 The greatest common factor = 7x<\/p>\n\n\n\n

4. 42x2<\/sup>yz and 63x3<\/sup>y2<\/sup>z3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 42 and 63.<\/p>\n\n\n\n

Greatest common factor of 42, 63 is 21.<\/p>\n\n\n\n

Common literals appearing in given numerical are x, y and z<\/p>\n\n\n\n

Smallest power of x in two monomials is 2<\/p>\n\n\n\n

Smallest power of y in two monomials is 1<\/p>\n\n\n\n

Smallest power of z in two monomials is 1<\/p>\n\n\n\n

Monomials of common literals with smallest power is x2<\/sup>yz<\/p>\n\n\n\n

\u2234 The greatest common factor = 21x2<\/sup>yz<\/p>\n\n\n\n

5. 12ax2<\/sup>, 6a2<\/sup>x3<\/sup> and 2a3<\/sup>x5<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 12, 6 and 2<\/p>\n\n\n\n

Greatest common factor of 12, 6 and 2 is 2.<\/p>\n\n\n\n

Common literals appearing in given numerical are a and x<\/p>\n\n\n\n

Smallest power of x in three monomials is 2<\/p>\n\n\n\n

Smallest power of a in three monomials is 1<\/p>\n\n\n\n

Monomials of common literals with smallest power is ax2<\/sup><\/p>\n\n\n\n

\u2234 The greatest common factor = 2ax2<\/sup><\/p>\n\n\n\n

6. 9x2<\/sup>, 15x2<\/sup>y3<\/sup>, 6xy2<\/sup> and 21x2<\/sup>y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 9, 15, 16 and 21<\/p>\n\n\n\n

Greatest common factor of 9, 15, 16 and 21 is 3.<\/p>\n\n\n\n

Common literals appearing in given numerical are x and y<\/p>\n\n\n\n

Smallest power of x in four monomials is 1<\/p>\n\n\n\n

Smallest power of y in four monomials is 0<\/p>\n\n\n\n

Monomials of common literals with smallest power is x<\/p>\n\n\n\n

\u2234 The greatest common factor = 3x<\/p>\n\n\n\n

7. 4a2<\/sup>b3<\/sup>, -12a3<\/sup>b, 18a4<\/sup>b3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 4, -12 and 18.<\/p>\n\n\n\n

Greatest common factor of 4, -12 and 18 is 2.<\/p>\n\n\n\n

Common literals appearing in given numerical are a and b<\/p>\n\n\n\n

Smallest power of a in three monomials is 2<\/p>\n\n\n\n

Smallest power of b in three monomials is 1<\/p>\n\n\n\n

Monomials of common literals with smallest power is a2<\/sup>b<\/p>\n\n\n\n

\u2234 The greatest common factor = 2a2<\/sup>b<\/p>\n\n\n\n

8. 6x2<\/sup>y2<\/sup>, 9xy3<\/sup>, 3x3<\/sup>y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 6, 9 and 3<\/p>\n\n\n\n

Greatest common factor of 6, 9 and 3 is 3.<\/p>\n\n\n\n

Common literals appearing in given numerical are x and y<\/p>\n\n\n\n

Smallest power of x in three monomials is 1<\/p>\n\n\n\n

Smallest power of y in three monomials is 2<\/p>\n\n\n\n

Monomials of common literals with smallest power is xy2<\/sup><\/p>\n\n\n\n

\u2234 The greatest common factor = 3xy2<\/sup><\/p>\n\n\n\n

9. a2<\/sup>b3<\/sup>, a3<\/sup>b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 0<\/p>\n\n\n\n

Common literals appearing in given numerical are a and b<\/p>\n\n\n\n

Smallest power of a in two monomials = 2<\/p>\n\n\n\n

Smallest power of b in two monomials = 2<\/p>\n\n\n\n

Monomials of common literals with smallest power is a2<\/sup>b2<\/sup><\/p>\n\n\n\n

\u2234 The greatest common factor = a2<\/sup>b2<\/sup><\/p>\n\n\n\n

10. 36a2<\/sup>b2<\/sup>c4<\/sup>, 54a5<\/sup>c2<\/sup>, 90a4<\/sup>b2<\/sup>c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 36, 54 and 90<\/p>\n\n\n\n

Greatest common factor of 36, 54 and 90 is 18.<\/p>\n\n\n\n

Common literals appearing in given numerical are a, b and c<\/p>\n\n\n\n

Smallest power of a in three monomials is 2<\/p>\n\n\n\n

Smallest power of b in three monomials is 0<\/p>\n\n\n\n

Smallest power of c in three monomials is 2<\/p>\n\n\n\n

Monomials of common literals with smallest power is a2<\/sup>c2<\/sup><\/p>\n\n\n\n

\u2234 The greatest common factor = 18a2<\/sup>c2<\/sup><\/p>\n\n\n\n

11. x3<\/sup>, -yx2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 0<\/p>\n\n\n\n

Common literals appearing in given numerical are x and y<\/p>\n\n\n\n

Smallest power of x in two monomials is 2<\/p>\n\n\n\n

Smallest power of y in two monomials is 0<\/p>\n\n\n\n

Monomials of common literals with smallest power is x2<\/sup><\/p>\n\n\n\n

\u2234 The greatest common factor = x2<\/sup><\/p>\n\n\n\n

12. 15a3<\/sup>, -45a2<\/sup>, -150a<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 15, -45 and 150<\/p>\n\n\n\n

Greatest common factor of 15, -45 and 150 is 15.<\/p>\n\n\n\n

Common literals appearing in given numerical is a<\/p>\n\n\n\n

Smallest power of a in three monomials is 1<\/p>\n\n\n\n

Monomials of common literals with smallest power is a<\/p>\n\n\n\n

\u2234 The greatest common factor = 15a<\/p>\n\n\n\n

13. 2x3<\/sup>y2<\/sup>, 10x2<\/sup>y3<\/sup>, 14xy<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 2, 10 and 14.<\/p>\n\n\n\n

Greatest common factor of 2, 10 and 14 is 2.<\/p>\n\n\n\n

Common literals appearing in given numerical are x and y<\/p>\n\n\n\n

Smallest power of x in three monomials is 1<\/p>\n\n\n\n

Smallest power of y in three monomials is 1<\/p>\n\n\n\n

Monomials of common literals with smallest power is xy<\/p>\n\n\n\n

\u2234 The greatest common factor = 2xy<\/p>\n\n\n\n

14. 14x3<\/sup>y5<\/sup>, 10x5<\/sup>y3<\/sup>, 2x2<\/sup>y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the numerical coefficients of given numerical are 14, 10 and 2.<\/p>\n\n\n\n

Greatest common factor of 14, 10 and 2 is 2.<\/p>\n\n\n\n

Common literals appearing in given numerical are x and y<\/p>\n\n\n\n

Smallest power of x in three monomials is 2<\/p>\n\n\n\n

Smallest power of y in three monomials is 2<\/p>\n\n\n\n

Monomials of common literals with smallest power is x2<\/sup>y2<\/sup><\/p>\n\n\n\n

\u2234 The greatest common factor = 2x2<\/sup>y2<\/sup><\/p>\n\n\n\n

Find the greatest common factor of the terms in each of the following expressions:<\/strong><\/p>\n\n\n\n

15. 5a4<\/sup> + 10a3<\/sup> \u2013 15a2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The greatest common factor of the three terms is 5a2<\/sup><\/p>\n\n\n\n

16. 2xyz + 3x2<\/sup>y + 4y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The greatest common factor of the three terms is y<\/p>\n\n\n\n

17. 3a2<\/sup>b2<\/sup> + 4b2<\/sup>c2<\/sup> + 12a2<\/sup>b2<\/sup>c2<\/sup><\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

The greatest common factor of the three terms is b2<\/sup>.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.2 PAGE NO: 7.5<\/h4>\n\n\n\n

Factorize the following:<\/strong><\/p>\n\n\n\n

1. 3x \u2013 9<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The greatest common factor in the given two terms is 3<\/p>\n\n\n\n

3x \u2013 9<\/p>\n\n\n\n

3 (x \u2013 3)<\/p>\n\n\n\n

2. 5x \u2013 15x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The greatest common factor in the given two terms is 5x<\/p>\n\n\n\n

5x \u2013 15x2<\/sup><\/p>\n\n\n\n

5x (1 \u2013 3x)<\/p>\n\n\n\n

3. 20a12<\/sup>b2<\/sup> \u2013 15a8<\/sup>b4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given two terms is 5a8<\/sup>b2<\/sup><\/p>\n\n\n\n

20a12<\/sup>b2<\/sup> \u2013 15a8<\/sup>b4<\/sup><\/p>\n\n\n\n

5a8<\/sup>b2<\/sup> (4a4<\/sup> \u2013 3b2<\/sup>)<\/p>\n\n\n\n

4. 72x6<\/sup>y7<\/sup> \u2013 96x7<\/sup>y6<\/sup>
Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given two terms is 24x6<\/sup>y6<\/sup><\/p>\n\n\n\n

72x6<\/sup>y7<\/sup> \u2013 96x7<\/sup>y6<\/sup><\/p>\n\n\n\n

24x6<\/sup>y6<\/sup> (3y \u2013 4x)<\/p>\n\n\n\n

5. 20x3<\/sup> \u2013 40x2<\/sup> + 80x<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is 20x<\/p>\n\n\n\n

20x3<\/sup> \u2013 40x2<\/sup> + 80x<\/p>\n\n\n\n

20x (x2<\/sup> \u2013 2x +4)<\/p>\n\n\n\n

6. 2x3<\/sup>y2<\/sup> \u2013 4x2<\/sup>y3<\/sup> + 8xy4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is 2xy2<\/sup><\/p>\n\n\n\n

2x3<\/sup>y2<\/sup> \u2013 4x2<\/sup>y3<\/sup> + 8xy4<\/sup><\/p>\n\n\n\n

2xy2<\/sup> (x2<\/sup> \u2013 2xy + 4y2<\/sup>)<\/p>\n\n\n\n

7. 10m3<\/sup>n2<\/sup> + 15m4<\/sup>n \u2013 20m2<\/sup>n3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is 5mn2<\/sup><\/p>\n\n\n\n

10m3<\/sup>n2<\/sup> + 15m4<\/sup>n \u2013 20m2<\/sup>n3<\/sup><\/p>\n\n\n\n

5m2<\/sup>n (2mn + 3m2<\/sup> \u2013 4n2<\/sup>)<\/p>\n\n\n\n

8. 2a4<\/sup>b4<\/sup> \u2013 3a3<\/sup>b5<\/sup> + 4a2<\/sup>b5<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is a2<\/sup>b4<\/sup><\/p>\n\n\n\n

2a4<\/sup>b4<\/sup> \u2013 3a3<\/sup>b5<\/sup> + 4a2<\/sup>b5<\/sup><\/p>\n\n\n\n

a2<\/sup>b4<\/sup> (2a2<\/sup> \u2013 3ab + 4b)<\/p>\n\n\n\n

9. 28a2<\/sup> + 14a2<\/sup>b2<\/sup> \u2013 21a4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is 7a2<\/sup><\/p>\n\n\n\n

28a2<\/sup> + 14a2<\/sup>b2<\/sup> \u2013 21a4<\/sup><\/p>\n\n\n\n

7a2 <\/sup>(4a + 2b2<\/sup> \u2013 3a2<\/sup>)<\/p>\n\n\n\n

10. a4<\/sup>b \u2013 3a2<\/sup>b2<\/sup> \u2013 6ab3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is ab<\/p>\n\n\n\n

a4<\/sup>b \u2013 3a2<\/sup>b2<\/sup> \u2013 6ab3<\/sup><\/p>\n\n\n\n

ab (a3<\/sup> \u2013 3ab \u2013 6b2<\/sup>)<\/p>\n\n\n\n

11. 2l2<\/sup>mn \u2013 3lm2<\/sup>n + 4lmn2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is lmn<\/p>\n\n\n\n

2l2<\/sup>mn \u2013 3lm2<\/sup>n + 4lmn2<\/sup><\/p>\n\n\n\n

lmn (2l \u2013 3m + 4n)<\/p>\n\n\n\n

12. x4<\/sup>y2<\/sup> \u2013 x2<\/sup>y4<\/sup> \u2013 x4<\/sup>y4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is x2<\/sup>y2<\/sup><\/p>\n\n\n\n

x4<\/sup>y2<\/sup> \u2013 x2<\/sup>y4<\/sup> \u2013 x4<\/sup>y4<\/sup><\/p>\n\n\n\n

x2<\/sup>y2<\/sup> (x2<\/sup> \u2013 y2<\/sup> \u2013 x2<\/sup>y2<\/sup>)<\/p>\n\n\n\n

13. 9x2<\/sup>y + 3axy<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is 3xy<\/p>\n\n\n\n

9x2<\/sup>y + 3axy<\/p>\n\n\n\n

3xy (3x + a)<\/p>\n\n\n\n

14. 16m \u2013 4m2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given two terms is 4m<\/p>\n\n\n\n

16m \u2013 4m2<\/sup><\/p>\n\n\n\n

4m (4 \u2013 m)<\/p>\n\n\n\n

15. -4a2<\/sup> + 4ab \u2013 4ca<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is \u2013 4a<\/p>\n\n\n\n

-4a2<\/sup> + 4ab \u2013 4ca<\/p>\n\n\n\n

-4a (a \u2013 b + c)<\/p>\n\n\n\n

16. <\/strong>x2<\/sup>yz + xy2<\/sup>z + xyz2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is xyz<\/p>\n\n\n\n

x2<\/sup>yz + xy2<\/sup>z + xyz2<\/sup><\/p>\n\n\n\n

xyz (x + y +z)<\/p>\n\n\n\n

17. ax2<\/sup>y + bxy2<\/sup> + cxyz<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Greatest common factor in the given three terms is xy<\/p>\n\n\n\n

ax2<\/sup>y + bxy2<\/sup> + cxyz<\/p>\n\n\n\n

xy (ax + by + cz)<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.3 PAGE NO: 7.7<\/h4>\n\n\n\n

Factorize each of the following algebraic expressions:<\/strong><\/p>\n\n\n\n

1. 6x (2x \u2013 y) + 7y (2x \u2013 y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

6x (2x \u2013 y) + 7y (2x \u2013 y)<\/p>\n\n\n\n

By taking (2x \u2013 y) as common we get,<\/p>\n\n\n\n

(6x + 7y) (2x \u2013 y)<\/p>\n\n\n\n

2. 2r (y \u2013 x) + s (x \u2013 y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

2r (y \u2013 x) + s (x \u2013 y)<\/p>\n\n\n\n

By taking (-1) as common we get,<\/p>\n\n\n\n

-2r (x \u2013 y) + s (x \u2013 y)<\/p>\n\n\n\n

By taking (x \u2013 y) as common we get,<\/p>\n\n\n\n

(x \u2013 y) (-2r + s)<\/p>\n\n\n\n

(x \u2013 y) (s \u2013 2r)<\/p>\n\n\n\n

3. 7a (2x \u2013 3) + 3b (2x \u2013 3)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

7a (2x \u2013 3) + 3b (2x \u2013 3)<\/p>\n\n\n\n

By taking (2x \u2013 3) as common we get,<\/p>\n\n\n\n

(7a + 3b) (2x \u2013 3)<\/p>\n\n\n\n

4. 9a (6a \u2013 5b) \u2013 12a2<\/sup> (6a \u2013 5b)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

9a (6a \u2013 5b) \u2013 12a2<\/sup> (6a \u2013 5b)<\/p>\n\n\n\n

By taking (6a \u2013 5b) as common we get,<\/p>\n\n\n\n

(9a \u2013 12a2<\/sup>) (6a \u2013 5b)<\/p>\n\n\n\n

3a(3 \u2013 4a) (6a \u2013 5b)<\/p>\n\n\n\n

5. 5 (x \u2013 2y)2<\/sup> + 3 (x \u2013 2y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

5 (x \u2013 2y)2<\/sup> + 3 (x \u2013 2y)<\/p>\n\n\n\n

By taking (x \u2013 2y) as common we get,<\/p>\n\n\n\n

(x \u2013 2y) [5 (x \u2013 2y) + 3]<\/p>\n\n\n\n

(x \u2013 2y) (5x \u2013 10y + 3)<\/p>\n\n\n\n

6. 16 (2l \u2013 3m)2<\/sup> \u2013 12 (3m \u2013 2l)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

16 (2l \u2013 3m)2<\/sup> \u2013 12 (3m \u2013 2l)<\/p>\n\n\n\n

By taking (-1) as common we get,<\/p>\n\n\n\n

16 (2l \u2013 3m)2<\/sup> + 12 (2l \u2013 3m)<\/p>\n\n\n\n

By taking 4(2l \u2013 3m) as common we get,<\/p>\n\n\n\n

4(2l \u2013 3m) [4 (2l \u2013 3m) + 3]<\/p>\n\n\n\n

4(2l \u2013 3m) (8l \u2013 12m + 3)<\/p>\n\n\n\n

7. 3a (x \u2013 2y) \u2013 b (x \u2013 2y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

3a (x \u2013 2y) \u2013 b (x \u2013 2y)<\/p>\n\n\n\n

By taking (x \u2013 2y) as common we get,<\/p>\n\n\n\n

(3a \u2013 b) (x \u2013 2y)<\/p>\n\n\n\n

8. a2<\/sup> (x + y) + b2<\/sup> (x + y) + c2<\/sup> (x + y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> (x + y) + b2<\/sup> (x + y) + c2<\/sup> (x + y)<\/p>\n\n\n\n

By taking (x + y) as common we get,<\/p>\n\n\n\n

(a2<\/sup> + b2<\/sup> + c2<\/sup>) (x + y)<\/p>\n\n\n\n

9. (x \u2013 y)2<\/sup> + (x \u2013 y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(x \u2013 y)2<\/sup> + (x \u2013 y)<\/p>\n\n\n\n

By taking (x \u2013 y) as common we get,<\/p>\n\n\n\n

(x \u2013 y) (x \u2013 y + 1)<\/p>\n\n\n\n

10. 6 (a + 2b) \u2013 4 (a + 2b)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

6 (a + 2b) \u2013 4 (a + 2b)2<\/sup><\/p>\n\n\n\n

By taking (a + 2b) as common we get,[6 \u2013 4 (a + 2b)] (a + 2b)<\/p>\n\n\n\n

(6 \u2013 4a \u2013 8b) (a + 2b)<\/p>\n\n\n\n

2(3 \u2013 2a \u2013 4b) (a + 2b)<\/p>\n\n\n\n

11. a (x \u2013 y) + 2b (y \u2013 x) + c (x \u2013 y)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a (x \u2013 y) + 2b (y \u2013 x) + c (x \u2013 y)2<\/sup><\/p>\n\n\n\n

By taking (-1) as common we get,<\/p>\n\n\n\n

a (x \u2013 y) \u2013 2b (x \u2013 y) + c (x \u2013 y)2<\/sup><\/p>\n\n\n\n

By taking (x \u2013 y) as common we get,[a \u2013 2b + c(x \u2013 y)] (x \u2013 y)<\/p>\n\n\n\n

(x \u2013 y) (a \u2013 2b + cx \u2013 cy)<\/p>\n\n\n\n

12. -4 (x \u2013 2y)2<\/sup> + 8 (x \u2013 2y)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

-4 (x \u2013 2y)2<\/sup> + 8 (x \u2013 2y)<\/p>\n\n\n\n

By taking 4(x \u2013 2y) as common we get,[-(x \u2013 2y) + 2] 4(x \u2013 2y)<\/p>\n\n\n\n

4(x \u2013 2y) (-x + 2y + 2)<\/p>\n\n\n\n

13. x3<\/sup> (a \u2013 2b) + x2<\/sup> (a \u2013 2b)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x3<\/sup> (a \u2013 2b) + x2<\/sup> (a \u2013 2b)<\/p>\n\n\n\n

By taking x2<\/sup> (a \u2013 2b) as common we get,<\/p>\n\n\n\n

(x + 1) [x2<\/sup> (a \u2013 2b)]<\/p>\n\n\n\n

x2<\/sup> (a \u2013 2b) (x + 1)<\/p>\n\n\n\n

14. (2x \u2013 3y) (a + b) + (3x \u2013 2y) (a + b)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(2x \u2013 3y) (a + b) + (3x \u2013 2y) (a + b)<\/p>\n\n\n\n

By taking (a + b) as common we get,<\/p>\n\n\n\n

(a + b) [(2x \u2013 3y) + (3x \u2013 2y)]<\/p>\n\n\n\n

(a + b) [2x -3y + 3x \u2013 2y]<\/p>\n\n\n\n

(a + b) [5x \u2013 5y]<\/p>\n\n\n\n

(a + b) 5(x \u2013 y)<\/p>\n\n\n\n

15. 4(x + y) (3a \u2013 b) + 6(x + y) (2b \u2013 3a)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

4(x + y) (3a \u2013 b) + 6(x + y) (2b \u2013 3a)<\/p>\n\n\n\n

By taking (x + y) as common we get,<\/p>\n\n\n\n

(x + y) [4(3a \u2013 b) + 6(2b \u2013 3a)]<\/p>\n\n\n\n

(x + y) [12a \u2013 4b + 12b \u2013 18a]<\/p>\n\n\n\n

(x + y) [-6a + 8b]<\/p>\n\n\n\n

(x + y) 2(-3a + 4b)<\/p>\n\n\n\n

(x + y) 2(4b \u2013 3a)<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.4 PAGE NO: 7.12<\/h4>\n\n\n\n

Factorize each of the following expressions:<\/strong><\/p>\n\n\n\n

1. qr \u2013 pr + qs \u2013 ps<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

qr \u2013 pr + qs \u2013 ps<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

qr + qs \u2013 pr \u2013 ps<\/p>\n\n\n\n

q(r + s) \u2013p (r + s)<\/p>\n\n\n\n

(q \u2013 p) (r + s)<\/p>\n\n\n\n

2. p2<\/sup>q \u2013 pr2<\/sup> \u2013 pq + r2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

p2<\/sup>q \u2013 pr2<\/sup> \u2013 pq + r2<\/sup><\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

p2<\/sup>q \u2013 pq \u2013 pr2<\/sup> + r2<\/sup><\/p>\n\n\n\n

pq(p \u2013 1) \u2013r2<\/sup>(p \u2013 1)<\/p>\n\n\n\n

(p \u2013 1) (pq \u2013 r2<\/sup>)<\/p>\n\n\n\n

3. 1 + x + xy + x2<\/sup>y<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

1 + x + xy + x2<\/sup>y<\/p>\n\n\n\n

1 (1 + x) + xy(1 + x)<\/p>\n\n\n\n

(1 + x) (1 + xy)<\/p>\n\n\n\n

4. ax + ay \u2013 bx \u2013 by<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

ax + ay \u2013 bx \u2013 by<\/p>\n\n\n\n

a(x + y) \u2013b (x + y)<\/p>\n\n\n\n

(a \u2013 b) (x + y)<\/p>\n\n\n\n

5. xa2<\/sup> + xb2<\/sup> \u2013 ya2<\/sup> \u2013 yb2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

xa2<\/sup> + xb2<\/sup> \u2013 ya2<\/sup> \u2013 yb2<\/sup><\/p>\n\n\n\n

x(a2<\/sup> + b2<\/sup>) \u2013y (a2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

(x \u2013 y) (a2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

6. x2<\/sup> + xy + xz + yz<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> + xy + xz + yz<\/p>\n\n\n\n

x (x + y) + z (x + y)<\/p>\n\n\n\n

(x + y) (x + z)<\/p>\n\n\n\n

7. 2ax + bx + 2ay + by<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

2ax + bx + 2ay + by<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

2ax + 2ay + bx + by<\/p>\n\n\n\n

2a (x + y) + b (x + y)<\/p>\n\n\n\n

(2a + b) (x + y)<\/p>\n\n\n\n

8. ab \u2013 by \u2013 ay + y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

ab \u2013 by \u2013 ay + y2<\/sup><\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

Ab \u2013 ay \u2013 by + y2<\/sup><\/p>\n\n\n\n

a (b \u2013 y) \u2013 y (b \u2013 y)<\/p>\n\n\n\n

(a \u2013 y) (b \u2013 y)<\/p>\n\n\n\n

9. axy + bcxy \u2013 az \u2013 bcz<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

axy + bcxy \u2013 az \u2013 bcz<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

axy \u2013 az + bcxy \u2013 bcz<\/p>\n\n\n\n

a (xy \u2013 z) + bc (xy \u2013 z)<\/p>\n\n\n\n

(a + bc) (xy \u2013 z)<\/p>\n\n\n\n

10. lm2<\/sup> \u2013 mn2<\/sup> \u2013 lm + n2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

lm2<\/sup> \u2013 mn2<\/sup> \u2013 lm + n2<\/sup><\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

lm2<\/sup> \u2013 lm \u2013 mn2<\/sup> + n2<\/sup><\/p>\n\n\n\n

lm (m \u2013 1) \u2013 n2<\/sup> (m \u2013 1)<\/p>\n\n\n\n

(lm \u2013 n2<\/sup>) (m \u2013 1)<\/p>\n\n\n\n

11. x3<\/sup> \u2013 y2<\/sup> + x \u2013 x2<\/sup>y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x3<\/sup> \u2013 y2<\/sup> + x \u2013 x2<\/sup>y2<\/sup><\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

x + x3<\/sup> \u2013 y2<\/sup> \u2013 x2<\/sup>y2<\/sup><\/p>\n\n\n\n

x (1 + x2<\/sup>) \u2013 y2<\/sup> (1 + x2<\/sup>)<\/p>\n\n\n\n

(x \u2013 y2<\/sup>) (1 + x2<\/sup>)<\/p>\n\n\n\n

12. 6xy + 6 \u2013 9y \u2013 4x<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

6xy + 6 \u2013 9y \u2013 4x<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

6xy \u2013 4x \u2013 9y + 6<\/p>\n\n\n\n

2x (3y \u2013 2) \u2013 3 (3y \u2013 2)<\/p>\n\n\n\n

(2x \u2013 3) (3y \u2013 2)<\/p>\n\n\n\n

13. x2<\/sup> \u2013 2ax \u2013 2ab + bx<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> \u2013 2ax \u2013 2ab + bx<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

x2<\/sup> + bx \u2013 2ax \u2013 2ab<\/p>\n\n\n\n

x (x + b) \u2013 2a (x + b)<\/p>\n\n\n\n

(x \u2013 2a) (x + b)<\/p>\n\n\n\n

14. x3<\/sup> \u2013 2x2<\/sup>y + 3xy2<\/sup> \u2013 6y3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x3<\/sup> \u2013 2x2<\/sup>y + 3xy2<\/sup> \u2013 6y3<\/sup><\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

x3<\/sup> + 3xy2<\/sup> \u2013 2x2<\/sup>y \u2013 6y3<\/sup><\/p>\n\n\n\n

x (x2<\/sup> + 3y2<\/sup>) \u2013 2y (x2<\/sup> + 3y2<\/sup>)<\/p>\n\n\n\n

(x \u2013 2y) (x2<\/sup> + 3y2<\/sup>)<\/p>\n\n\n\n

15. abx2<\/sup> + (ay \u2013 b) x \u2013 y<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

abx2<\/sup> + (ay \u2013 b) x \u2013 y<\/p>\n\n\n\n

abx2<\/sup> + ayx \u2013 bx \u2013 y<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

abx2<\/sup> \u2013 bx + ayx \u2013 y<\/p>\n\n\n\n

bx (ax \u2013 1) + y (ax \u2013 1)<\/p>\n\n\n\n

(bx + y) (ax \u2013 1)<\/p>\n\n\n\n

16. (ax + by)2<\/sup> + (bx \u2013 ay)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(ax + by)2<\/sup> + (bx \u2013 ay)2<\/sup><\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + b2<\/sup>y2<\/sup> + 2axby + b2<\/sup>x2<\/sup> + a2<\/sup>y2<\/sup> \u2013 2axby<\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + b2<\/sup>y2<\/sup> + b2<\/sup>x2<\/sup> + a2<\/sup>y2<\/sup><\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + a2<\/sup>y2<\/sup> + b2<\/sup>y2<\/sup> + b2<\/sup>x2<\/sup><\/p>\n\n\n\n

a2<\/sup> (x2<\/sup> + y2<\/sup>) + b2<\/sup> (x2<\/sup> + y2<\/sup>)<\/p>\n\n\n\n

(a2<\/sup> + b2<\/sup>) (x2<\/sup> + y2<\/sup>)<\/p>\n\n\n\n

17. 16 (a \u2013 b)3<\/sup> \u2013 24 (a \u2013 b)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

16(a \u2013 b)3<\/sup> \u2013 24(a \u2013 b)2<\/sup><\/p>\n\n\n\n

8 (a \u2013 b)2<\/sup> [2 (a \u2013 b) \u2013 3]<\/p>\n\n\n\n

8 (a \u2013 b)2<\/sup> (2a \u2013 2b \u2013 3)<\/p>\n\n\n\n

18. ab (x2<\/sup> + 1) + x (a2<\/sup> + b2<\/sup>)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

ab(x2<\/sup> + 1) + x(a2<\/sup> + b2<\/sup>)<\/p>\n\n\n\n

abx2<\/sup> + ab + xa2<\/sup> + xb2<\/sup><\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

abx2<\/sup> + xa2<\/sup> + xb2<\/sup> + ab<\/p>\n\n\n\n

ax (bx + a) + b (bx + a)<\/p>\n\n\n\n

(ax + b) (bx + a)<\/p>\n\n\n\n

19. a2<\/sup>x2<\/sup> + (ax2<\/sup> + 1)x + a<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + (ax2<\/sup> + 1)x + a<\/p>\n\n\n\n

a2<\/sup>x2<\/sup> + ax3<\/sup> + x + a<\/p>\n\n\n\n

ax2<\/sup> (a + x) + 1 (x + a)<\/p>\n\n\n\n

(x + a) (ax2<\/sup> + 1)<\/p>\n\n\n\n

20. a (a \u2013 2b \u2013 c) + 2bc<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a (a \u2013 2b \u2013 c) + 2bc<\/p>\n\n\n\n

a2<\/sup> \u2013 2ab \u2013 ac + 2bc<\/p>\n\n\n\n

a (a \u2013 2b) \u2013 c (a \u2013 2b)<\/p>\n\n\n\n

(a \u2013 2b) (a \u2013 c)<\/p>\n\n\n\n

21. a (a + b \u2013 c) \u2013 bc<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a (a + b \u2013 c) \u2013 bc<\/p>\n\n\n\n

a2<\/sup> + ab \u2013 ac \u2013 bc<\/p>\n\n\n\n

a (a + b) \u2013 c (a + b)<\/p>\n\n\n\n

(a + b) (a \u2013 c)<\/p>\n\n\n\n

22. x2<\/sup> \u2013 11xy \u2013 x + 11y<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> \u2013 11xy \u2013 x + 11y<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

x2<\/sup> \u2013 x \u2013 11xy + 11y<\/p>\n\n\n\n

x (x \u2013 1) \u2013 11y (x \u2013 1)<\/p>\n\n\n\n

(x \u2013 11y) (x \u2013 1)<\/p>\n\n\n\n

23. ab \u2013 a \u2013 b + 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

ab \u2013 a \u2013 b + 1<\/p>\n\n\n\n

a (b \u2013 1) \u2013 1 (b \u2013 1)<\/p>\n\n\n\n

(a \u2013 1) (b \u2013 1)<\/p>\n\n\n\n

24. x2<\/sup> + y \u2013 xy \u2013 x<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> + y \u2013 xy \u2013 x<\/p>\n\n\n\n

By grouping similar terms we get,<\/p>\n\n\n\n

x2<\/sup> \u2013 x + y \u2013 xy<\/p>\n\n\n\n

x (x \u2013 1) \u2013 y (x \u2013 1)<\/p>\n\n\n\n

(x \u2013 y) (x \u2013 1)<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.5 PAGE NO: 7.17<\/h4>\n\n\n\n

Factorize each of the following expressions:<\/strong><\/p>\n\n\n\n

1. 16x2<\/sup> \u2013 25y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

16x2<\/sup> \u2013 25y2<\/sup><\/p>\n\n\n\n

(4x)2<\/sup> \u2013 (5y)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b) we get,<\/p>\n\n\n\n

(4x + 5y) (4x \u2013 5y)<\/p>\n\n\n\n

2. 27x2<\/sup> \u2013 12y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

27x2<\/sup> \u2013 12y2<\/sup><\/p>\n\n\n\n

By taking 3 as common we get,<\/p>\n\n\n\n

3 [(3x)2<\/sup> \u2013 (2y)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

3 (3x + 2y) (3x \u2013 2y)<\/p>\n\n\n\n

3. 144a2<\/sup> \u2013 289b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

144a2<\/sup> \u2013 289b2<\/sup><\/p>\n\n\n\n

(12a)2<\/sup> \u2013 (17b)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(12a + 17b) (12a \u2013 17b)<\/p>\n\n\n\n

4. 12m2<\/sup> \u2013 27<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

12m2<\/sup> \u2013 27<\/p>\n\n\n\n

By taking 3 as common we get,<\/p>\n\n\n\n

3 (4m2<\/sup> \u2013 9)<\/p>\n\n\n\n

3 [(2m)2<\/sup> \u2013 32<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

3 (2m + 3) (2m \u2013 3)<\/p>\n\n\n\n

5. 125x2<\/sup> \u2013 45y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

125x2<\/sup> \u2013 45y2<\/sup><\/p>\n\n\n\n

By taking 5 as common we get,<\/p>\n\n\n\n

5 (25x2<\/sup> \u2013 9y2<\/sup>)<\/p>\n\n\n\n

5 [(5x)2<\/sup> \u2013 (3y)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

5 (5x + 3y) (5x \u2013 3y)<\/p>\n\n\n\n

6. 144a2<\/sup> \u2013 169b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

144a2<\/sup> \u2013 169b2<\/sup><\/p>\n\n\n\n

(12a)2<\/sup> \u2013 (13b)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(12a + 13b) (12a \u2013 13b)<\/p>\n\n\n\n

7. (2a \u2013 b)2<\/sup> \u2013 16c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(2a \u2013 b)2<\/sup> \u2013 16c2<\/sup><\/p>\n\n\n\n

(2a \u2013 b)2<\/sup> \u2013 (4c)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(2a \u2013 b + 4c) (2a \u2013 b \u2013 4c)<\/p>\n\n\n\n

8. (x + 2y)2<\/sup> \u2013 4 (2x \u2013 y)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(x + 2y)2<\/sup> \u2013 4 (2x \u2013 y)2<\/sup><\/p>\n\n\n\n

(x + 2y)2<\/sup> \u2013 [2 (2x \u2013 y)]2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>)= (a + b) (a \u2013 b) we get,[(x + 2y) + 2 (2x \u2013 y)] [x + 2y \u2013 2 (2x \u2013 y)]<\/p>\n\n\n\n

(x + 4x + 2y \u2013 2y) (x \u2013 4x + 2y + 2y)<\/p>\n\n\n\n

(5x) (4y \u2013 3x)<\/p>\n\n\n\n

9. 3a5<\/sup> \u2013 48a3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

3a5<\/sup> \u2013 48a3<\/sup><\/p>\n\n\n\n

By taking 3 as common we get,<\/p>\n\n\n\n

3a3<\/sup> (a2<\/sup> \u2013 16)<\/p>\n\n\n\n

3a3<\/sup> (a2<\/sup> \u2013 42<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

3a3<\/sup> (a + 4) (a \u2013 4)<\/p>\n\n\n\n

10. a4<\/sup> \u2013 16b4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a4<\/sup> \u2013 16b4<\/sup><\/p>\n\n\n\n

(a2<\/sup>)2<\/sup> \u2013 (4b2<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup> + 4b2<\/sup>) (a2<\/sup> \u2013 4b2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup> + 4b2<\/sup>) (a + 2b) (a \u2013 2b)<\/p>\n\n\n\n

11. x8<\/sup> \u2013 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x8<\/sup> \u2013 1<\/p>\n\n\n\n

(x4<\/sup>)2<\/sup>\u2013(1)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x4<\/sup> + 1) (x4<\/sup> \u2013 1)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x4<\/sup> + 1) (x2<\/sup> + 1) (x \u2013 1) (x + 1)<\/p>\n\n\n\n

12. 64 \u2013 (a + 1)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

64 \u2013 (a + 1)2<\/sup><\/p>\n\n\n\n

82<\/sup> \u2013 (a + 1)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[8 + (a + 1)] [8 \u2013 (a + 1)]<\/p>\n\n\n\n

(a + 9) (7 \u2013 a)<\/p>\n\n\n\n

13. 36l2<\/sup> \u2013 (m + n)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

36l2<\/sup> \u2013 (m + n)2<\/sup><\/p>\n\n\n\n

(6l)2<\/sup> \u2013 (m + n)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(6l + m + n) (6l \u2013 m \u2013 n)<\/p>\n\n\n\n

14. 25x4<\/sup>y4<\/sup> \u2013 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

25x4<\/sup>y4<\/sup> \u2013 1<\/p>\n\n\n\n

(5x2<\/sup>y2<\/sup>)2<\/sup> \u2013 (1)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(5x2<\/sup>y2<\/sup> \u2013 1) (5x2<\/sup>y2<\/sup> + 1)<\/p>\n\n\n\n

15. a4<\/sup> \u2013 1\/b4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a4<\/sup> \u2013 1\/b4<\/sup><\/p>\n\n\n\n

(a2<\/sup>)2<\/sup> \u2013 (1\/b2<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup> + 1\/b2<\/sup>) (a2<\/sup> \u2013 1\/b2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup> + 1\/b2<\/sup>) (a \u2013 1\/b) (a + 1\/b)<\/p>\n\n\n\n

16. x3<\/sup> \u2013 144x<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x3<\/sup> \u2013 144x<\/p>\n\n\n\n

x [x2<\/sup> \u2013 (12)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

x (x + 12) (x \u2013 12)<\/p>\n\n\n\n

17. (x \u2013 4y)2<\/sup> \u2013 625<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(x \u2013 4y)2<\/sup> \u2013 625<\/p>\n\n\n\n

(x \u2013 4y)2<\/sup> \u2013 (25)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x \u2013 4y + 25) (x \u2013 4y \u2013 25)<\/p>\n\n\n\n

18. 9 (a \u2013 b)2<\/sup> \u2013 100 (x \u2013 y)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

9 (a \u2013 b)2<\/sup> \u2013 100 (x \u2013 y)2<\/sup>[3 (a \u2013 b)]2<\/sup> \u2013 [10 (x \u2013 y)]2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[3 (a \u2013 b) + 10 (x + y)] [3 (a \u2013 b) \u2013 10 (x \u2013 y)] [3a \u2013 3b + 10x \u2013 10y] [3a \u2013 3b \u2013 10x + 10y]<\/p>\n\n\n\n

19. (3 + 2a)2<\/sup> \u2013 25a2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(3 + 2a)2<\/sup> \u2013 25a2<\/sup><\/p>\n\n\n\n

(3 + 2a)2<\/sup> \u2013 (5a)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(3 + 2a + 5a) (3 + 2a \u2013 5a)<\/p>\n\n\n\n

(3 + 7a) (3 \u2013 3a)<\/p>\n\n\n\n

(3 + 7a) 3(1 \u2013 a)<\/p>\n\n\n\n

20. (x + y)2<\/sup> \u2013 (a \u2013 b)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(x + y)2<\/sup> \u2013 (a \u2013 b)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[(x + y) + (a \u2013 b)] [(x + y) \u2013 (a \u2013 b)]<\/p>\n\n\n\n

(x + y + a \u2013 b) (x + y \u2013 a + b)<\/p>\n\n\n\n

21. 1\/16x2<\/sup>y2<\/sup> \u2013 4\/49y2<\/sup>z2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

1\/16x2<\/sup>y2<\/sup> \u2013 4\/49y2<\/sup>z2<\/sup><\/p>\n\n\n\n

(1\/4xy)2<\/sup> \u2013 (2\/7yz)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(xy\/4 + 2yz\/7) (xy\/4 \u2013 2yz\/7)<\/p>\n\n\n\n

y2<\/sup> (x\/4 + 2\/7z) (x\/4 \u2013 2\/7z)<\/p>\n\n\n\n

22. 75a3<\/sup>b2<\/sup> \u2013 108ab4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

75a3<\/sup>b2<\/sup> \u2013 108ab4<\/sup><\/p>\n\n\n\n

3ab2<\/sup> (25a2<\/sup> \u2013 36b2<\/sup>)<\/p>\n\n\n\n

3ab2<\/sup> [(5a)2<\/sup> \u2013 (6b)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

3ab2<\/sup> (5a + 6b) (5a \u2013 6b)<\/p>\n\n\n\n

23. x5<\/sup> \u2013 16x3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x5<\/sup> \u2013 16x3<\/sup><\/p>\n\n\n\n

x3<\/sup> (x2<\/sup> \u2013 16)<\/p>\n\n\n\n

x3<\/sup> (x2<\/sup> \u2013 42<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

x3<\/sup> (x + 4) (x \u2013 4)<\/p>\n\n\n\n

24. 50\/x2<\/sup> \u2013 2x2<\/sup>\/81<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

50\/x2<\/sup> \u2013 2x2<\/sup>\/81<\/p>\n\n\n\n

2 (25\/x2<\/sup> \u2013 x2<\/sup>\/81)<\/p>\n\n\n\n

2 [(5\/x)2<\/sup> \u2013 (x\/9)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

2 (5\/x+ x\/9) (5\/x \u2013 x\/9)<\/p>\n\n\n\n

25. 256x3<\/sup> \u2013 81x<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

256x3<\/sup> \u2013 81x<\/p>\n\n\n\n

x (256x4<\/sup> \u2013 81)<\/p>\n\n\n\n

x [(16x2<\/sup>)2<\/sup> \u2013 92<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

x (4x + 3) (4x \u2013 3) (16x2<\/sup> + 9)<\/p>\n\n\n\n

26. a4<\/sup> \u2013 (2b + c)4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a4<\/sup> \u2013 (2b + c)4<\/sup><\/p>\n\n\n\n

(a2<\/sup>)2<\/sup> \u2013 [(2b + c)2<\/sup>]2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[a2<\/sup> + (2b + c)2<\/sup>] [a2<\/sup> \u2013 (2b + c)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[a2<\/sup> + (2b + c)2<\/sup>] [a + 2b + c] [a \u2013 2b \u2013 c]<\/p>\n\n\n\n

27. (3x + 4y)4<\/sup> \u2013 x4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(3x + 4y)4<\/sup> \u2013 x4<\/sup>[(3x + 4y)2<\/sup>]2<\/sup> \u2013 (x2<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[(3x + 4y)2<\/sup> + x2<\/sup>] [(3x + 4y)2<\/sup> \u2013 x2<\/sup>] [(3x + 4y)2<\/sup> + x2<\/sup>] [3x + 4y + x] [3x + 4y \u2013 x] [(3x + 4y)2<\/sup> + x2<\/sup>] [4x + 4y] [2x + 4y] [(3x + 4y)2<\/sup> + x2<\/sup>] 8[x + 2y] [x + y]<\/p>\n\n\n\n

28. p2<\/sup>q2<\/sup> \u2013 p4<\/sup>q4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

p2<\/sup>q2<\/sup> \u2013 p4<\/sup>q4<\/sup><\/p>\n\n\n\n

(pq)2<\/sup> \u2013 (p2<\/sup>q2<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(pq + p2<\/sup>q2<\/sup>) (pq \u2013 p2<\/sup>q2<\/sup>)<\/p>\n\n\n\n

p2<\/sup>q2<\/sup> (1 + pq) (1 \u2013 pq)<\/p>\n\n\n\n

29. 3x3<\/sup>y \u2013 243xy3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

3x3<\/sup>y \u2013 243xy3<\/sup><\/p>\n\n\n\n

3xy (x2<\/sup> \u2013 81y2<\/sup>)<\/p>\n\n\n\n

3xy [x2<\/sup> \u2013 (9y)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(3xy) (x + 9y) (x \u2013 9y)<\/p>\n\n\n\n

30. a4<\/sup>b4<\/sup> \u2013 16c4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a4<\/sup>b4<\/sup> \u2013 16c4<\/sup><\/p>\n\n\n\n

(a2<\/sup>b2<\/sup>)2<\/sup> \u2013 (4c2<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup>b2<\/sup> + 4c2<\/sup>) (a2<\/sup>b2<\/sup> \u2013 4c2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup>b2<\/sup> + 4c2<\/sup>) (ab + 2c) (ab \u2013 2c)<\/p>\n\n\n\n

31. x4<\/sup> \u2013 625<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x4<\/sup> \u2013 625<\/p>\n\n\n\n

(x2<\/sup>)2<\/sup> \u2013 (25)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x2<\/sup> + 25) (x2<\/sup> \u2013 25)<\/p>\n\n\n\n

(x2<\/sup> + 25) (x2<\/sup> \u2013 52<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x2<\/sup> + 25) (x + 5) (x \u2013 5)<\/p>\n\n\n\n

32. x4<\/sup> \u2013 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x4<\/sup> \u2013 1<\/p>\n\n\n\n

(x2<\/sup>)2<\/sup> \u2013 (1)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x2<\/sup> + 1) (x2<\/sup> \u2013 1)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x2<\/sup> + 1) (x + 1) (x \u2013 1)<\/p>\n\n\n\n

33. 49(a \u2013 b)2<\/sup> \u2013 25(a + b)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

49(a \u2013 b)2<\/sup> \u2013 25(a + b)2<\/sup>[7 (a \u2013 b)]2<\/sup> \u2013 [5 (a + b)]2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[7 (a \u2013 b) + 5 (a + b)] [7 (a \u2013 b) \u2013 5 (a + b)]<\/p>\n\n\n\n

(7a \u2013 7b + 5a + 5b) (7a \u2013 7b \u2013 5a \u2013 5b)<\/p>\n\n\n\n

(12a \u2013 2b) (2a \u2013 12b)<\/p>\n\n\n\n

2 (6a \u2013 b) 2 (a \u2013 6b)<\/p>\n\n\n\n

4 (6a \u2013 b) (a \u2013 6b)<\/p>\n\n\n\n

34. x \u2013 y \u2013 x2<\/sup> + y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x \u2013 y \u2013 x2<\/sup> + y2<\/sup><\/p>\n\n\n\n

x \u2013 y \u2013 (x2<\/sup> \u2013 y2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

x \u2013 y \u2013 (x + y) (x \u2013 y)<\/p>\n\n\n\n

(x \u2013 y) (1 \u2013 x \u2013 y)<\/p>\n\n\n\n

35. 16(2x \u2013 1)2<\/sup> \u2013 25y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

16(2x \u2013 1)2<\/sup> \u2013 25y2<\/sup>[4 (2x \u2013 1)]2<\/sup> \u2013 (5y)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(8x + 5y \u2013 4) (8x \u2013 5y \u2013 4)<\/p>\n\n\n\n

36. 4(xy + 1)2<\/sup> \u2013 9(x \u2013 1)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

4(xy + 1)2<\/sup> \u2013 9(x \u2013 1)2<\/sup>[2 (xy + 1)]2<\/sup> \u2013 [3 (x \u2013 1)]2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(2xy + 2 + 3x \u2013 3) (2xy + 2 \u2013 3x + 3)<\/p>\n\n\n\n

(2xy + 3x \u2013 1) (2xy \u2013 3x + 5)<\/p>\n\n\n\n

37. (2x + 1)2<\/sup> \u2013 9x4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(2x + 1)2<\/sup> \u2013 9x4<\/sup><\/p>\n\n\n\n

(2x + 1)2<\/sup> \u2013 (3x2<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(2x + 1 + 3x2<\/sup>) (2x + 1 \u2013 3x2<\/sup>)<\/p>\n\n\n\n

(3x2<\/sup> + 2x + 1) (-3x2<\/sup> + 2x + 1)<\/p>\n\n\n\n

38. x4<\/sup> \u2013 (2y \u2013 3z)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x4<\/sup> \u2013 (2y \u2013 3z)2<\/sup><\/p>\n\n\n\n

(x2<\/sup>)2<\/sup> \u2013 (2y \u2013 3z)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(x2<\/sup> + 2y \u2013 3z) (x2<\/sup> \u2013 2y + 3z)<\/p>\n\n\n\n

39. a2 <\/sup>\u2013 b2<\/sup> + a \u2013 b<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> \u2013 b2<\/sup> + a \u2013 b<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a + b) (a \u2013 b) + (a \u2013 b)<\/p>\n\n\n\n

(a \u2013 b) (a + b + 1)<\/p>\n\n\n\n

40. 16a4<\/sup> \u2013 b4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

16a4<\/sup> \u2013 b4<\/sup><\/p>\n\n\n\n

(4a2<\/sup>)2<\/sup> \u2013 (b2<\/sup>)2<\/sup><\/p>\n\n\n\n

(4a2<\/sup> + b2<\/sup>) (4a2<\/sup> \u2013 b2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(4a2<\/sup> + b2<\/sup>) (2a + b) (2a \u2013 b)<\/p>\n\n\n\n

41. a4<\/sup> \u2013 16(b \u2013 c)4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a4<\/sup> \u2013 16(b \u2013 c)4<\/sup><\/p>\n\n\n\n

(a2<\/sup>)2<\/sup> \u2013 [4 (b \u2013 c)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[a2<\/sup> + 4 (b \u2013 c)2<\/sup>] [a2<\/sup> \u2013 4 (b \u2013 c)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)[a2<\/sup> + 4 (b \u2013 c)2<\/sup>] [(a + 2b \u2013 2c) (a \u2013 2b + 2c)]<\/p>\n\n\n\n

42. 2a5<\/sup> \u2013 32a<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

2a5<\/sup> \u2013 32a<\/p>\n\n\n\n

2a (a4<\/sup> \u2013 16)<\/p>\n\n\n\n

2a [(a2<\/sup>)2<\/sup> \u2013 (4)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

2a (a2<\/sup> + 4) (a2<\/sup> \u2013 4)<\/p>\n\n\n\n

2a (a2<\/sup> + 4) (a2<\/sup> \u2013 22<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

2a (a2<\/sup> + 4) (a + 2) (a \u2013 2)<\/p>\n\n\n\n

43. a4<\/sup>b4<\/sup> \u2013 81c4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a4<\/sup>b4<\/sup> \u2013 81c4<\/sup><\/p>\n\n\n\n

(a2<\/sup>b2<\/sup>)2<\/sup> \u2013 (9c2<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup>b2<\/sup> + 9c2<\/sup>) (a2<\/sup>b2<\/sup> \u2013 9c2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

(a2<\/sup>b2<\/sup> + 9c2<\/sup>) (ab + 3c) (ab \u2013 3c)<\/p>\n\n\n\n

44. xy9<\/sup> \u2013 yx9<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

xy9<\/sup> \u2013 yx9<\/sup><\/p>\n\n\n\n

-xy (x8<\/sup> \u2013 y8<\/sup>)<\/p>\n\n\n\n

-xy [(x4<\/sup>)2<\/sup> \u2013 (y4<\/sup>)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

-xy (x4<\/sup> + y4<\/sup>) (x4<\/sup> \u2013 y4<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

-xy (x4<\/sup> + y4<\/sup>) (x2<\/sup> + y2<\/sup>) (x2<\/sup> \u2013 y2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

-xy (x4<\/sup> + y4<\/sup>) (x2<\/sup> + y2<\/sup>) (x + y) (x \u2013 y)<\/p>\n\n\n\n

45. x3<\/sup> \u2013 x<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x3<\/sup> \u2013 x<\/p>\n\n\n\n

x (x2<\/sup> \u2013 1)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

x (x + 1) (x \u2013 1)<\/p>\n\n\n\n

46. 18a2<\/sup>x2<\/sup> \u2013 32<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

18a2<\/sup>x2<\/sup> \u2013 32<\/p>\n\n\n\n

2 [(3ax)2<\/sup> \u2013 (4)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a-b) (a+b)<\/p>\n\n\n\n

2 (3ax + 4) (3ax \u2013 4)<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.6 PAGE NO: 7.22<\/h3>\n\n\n\n

Factorize each of the following algebraic expressions:<\/strong><\/p>\n\n\n\n

1. 4x2<\/sup> + 12xy + 9y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

4x2<\/sup> + 12xy + 9y2<\/sup><\/p>\n\n\n\n

By using the formula (x + y)2<\/sup> = x2<\/sup> + y2<\/sup> + 2xy<\/p>\n\n\n\n

(2x)2<\/sup> + (3y)2<\/sup> + 2 (2x) (3y)<\/p>\n\n\n\n

(2x + 3y)2<\/sup><\/p>\n\n\n\n

(2x + 3y) (2x + 3y)<\/p>\n\n\n\n

2. 9a2<\/sup> \u2013 24ab + 16b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

9a2<\/sup> \u2013 24ab + 16b2<\/sup><\/p>\n\n\n\n

By using the formula (x \u2013 y)2<\/sup> = x2<\/sup> + y2<\/sup> \u2013 2xy<\/p>\n\n\n\n

Here x = 3a, y = 4b So,<\/p>\n\n\n\n

(3a)2<\/sup> + (4b)2<\/sup> \u2013 2 (3a) (4b)<\/p>\n\n\n\n

(3a \u2013 4b)2<\/sup><\/p>\n\n\n\n

(3a \u2013 4b) (3a \u2013 4b)<\/p>\n\n\n\n

3. p2<\/sup>q2<\/sup> \u2013 6pqr + 9r2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

p2<\/sup>q2<\/sup> \u2013 6pqr + 9r2<\/sup><\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(pq)2<\/sup> + (3r)2<\/sup> \u2013 2 (pq) (3r)<\/p>\n\n\n\n

(pq \u2013 3r)2<\/sup><\/p>\n\n\n\n

(pq \u2013 3r) (pq \u2013 3r)<\/p>\n\n\n\n

4. 36a2<\/sup> + 36a + 9<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

36a2<\/sup> + 36a + 9<\/p>\n\n\n\n

(6a)2<\/sup> + 2 \u00d7 6a \u00d7 3 + 32<\/sup><\/p>\n\n\n\n

(6a + 3)2<\/sup><\/p>\n\n\n\n

5. a2<\/sup> + 2ab + b2<\/sup> \u2013 16<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 2ab + b2<\/sup> \u2013 16<\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(a + b)2<\/sup> \u2013 42<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a + b + 4) (a + b \u2013 4)<\/p>\n\n\n\n

6. 9z2<\/sup> \u2013 x2<\/sup> + 4xy \u2013 4y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

9z2<\/sup> \u2013 x2<\/sup> + 4xy \u2013 4y2<\/sup><\/p>\n\n\n\n

(3z)2<\/sup> \u2013 [x2<\/sup> \u2013 2 (x) (2y) + (2y)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(3z)2<\/sup> \u2013 (x \u2013 2y)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)[(x \u2013 2y) + 3z] [\u2013x + 2y + 3z)]<\/p>\n\n\n\n

7. 9a4<\/sup> \u2013 24a2<\/sup>b2<\/sup> + 16b4<\/sup> \u2013 256<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

9a4<\/sup> \u2013 24a2<\/sup>b2<\/sup> + 16b4<\/sup> \u2013 256<\/p>\n\n\n\n

(3a2<\/sup>)2<\/sup> \u2013 2 (4a2<\/sup>) (3b2<\/sup>) + (4b2<\/sup>)2<\/sup> \u2013 (16)2<\/sup><\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(3a2<\/sup> \u2013 4b2<\/sup>)2<\/sup> \u2013 (16)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(3a2<\/sup> \u2013 4b2<\/sup> + 16) (3a2<\/sup> \u2013 4b2<\/sup> \u2013 16)<\/p>\n\n\n\n

8. 16 \u2013 a6<\/sup> + 4a3<\/sup>b3<\/sup> \u2013 4b6<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

16 \u2013 a6<\/sup> + 4a3<\/sup>b3<\/sup> \u2013 4b6<\/sup><\/p>\n\n\n\n

42<\/sup> \u2013 [(a3<\/sup>)2<\/sup> \u2013 2 (a3<\/sup>) (2b3<\/sup>) + (2b3<\/sup>)2<\/sup>]<\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

42<\/sup> \u2013 (a3<\/sup> \u2013 2b3<\/sup>)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)[4 + (a3<\/sup> \u2013 2b3<\/sup>)] [4 \u2013 (a3<\/sup> \u2013 2b3<\/sup>)]<\/p>\n\n\n\n

9. a2<\/sup> \u2013 2ab + b2<\/sup> \u2013 c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> \u2013 2ab + b2<\/sup> \u2013 c2<\/sup><\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(a \u2013 b)2<\/sup> \u2013 c2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a \u2013 b + c) (a \u2013 b \u2013 c)<\/p>\n\n\n\n

10. x2<\/sup> + 2x + 1 \u2013 9y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> + 2x + 1 \u2013 9y2<\/sup><\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(x + 1)2<\/sup> \u2013 (3y)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(x + 3y + 1) (x \u2013 3y + 1)<\/p>\n\n\n\n

11. a2<\/sup> + 4ab + 3b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 4ab + 3b2<\/sup><\/p>\n\n\n\n

By using factors for 3 i.e., 3 and 1<\/p>\n\n\n\n

a2<\/sup> + ab + 3ab + 3b2<\/sup><\/p>\n\n\n\n

By grouping we get,<\/p>\n\n\n\n

a (a + b) + 3b (a + b)<\/p>\n\n\n\n

(a + 3b) (a + b)<\/p>\n\n\n\n

12. 96 \u2013 4x \u2013 x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

96 \u2013 4x \u2013 x2<\/sup><\/p>\n\n\n\n

-x2<\/sup> \u2013 4x + 96<\/p>\n\n\n\n

By using factors for 96 i.e., 12 and 8<\/p>\n\n\n\n

-x2<\/sup> \u2013 12x + 8x + 96<\/p>\n\n\n\n

By grouping we get,<\/p>\n\n\n\n

-x (x + 12) + 8 (x + 12)<\/p>\n\n\n\n

(x + 12) (-x + 8)<\/p>\n\n\n\n

13. a4<\/sup> + 3a2<\/sup> + 4<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a4<\/sup> + 3a2<\/sup> + 4<\/p>\n\n\n\n

(a2<\/sup>)2<\/sup> + (a2<\/sup>)2<\/sup> + 2 (2a2<\/sup>) + 4 \u2013 a2<\/sup><\/p>\n\n\n\n

(a2<\/sup> + 2)2<\/sup> + (-a2<\/sup>)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a2<\/sup> + 2 + a) (a2<\/sup> + 2 \u2013 a)<\/p>\n\n\n\n

(a2<\/sup> + a + 2) (a2<\/sup> \u2013 a + 2)<\/p>\n\n\n\n

14. 4x4<\/sup> + 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

4x4<\/sup> + 1<\/p>\n\n\n\n

(2x2<\/sup>)2<\/sup> + 1 + 4x2<\/sup> \u2013 4x2<\/sup><\/p>\n\n\n\n

(2x2<\/sup> + 1)2<\/sup> \u2013 4x2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(2x2<\/sup> + 1 + 2x) (2x2<\/sup> + 1 \u2013 2x)<\/p>\n\n\n\n

(2x2<\/sup> + 2x + 1) (2x2<\/sup> \u2013 2x + 1)<\/p>\n\n\n\n

15. 4x4<\/sup> + y4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

4x4<\/sup> + y4<\/sup><\/p>\n\n\n\n

(2x2<\/sup>)2<\/sup> + (y2<\/sup>)2<\/sup> + 4x2<\/sup>y2<\/sup> \u2013 4x2<\/sup>y2<\/sup><\/p>\n\n\n\n

(2x2<\/sup> + y2<\/sup>)2<\/sup> \u2013 4x2<\/sup>y2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(2x2<\/sup> + y2<\/sup> + 2xy) (2x2<\/sup> + y2<\/sup> \u2013 2xy)<\/p>\n\n\n\n

16. (x + 2)4<\/sup> \u2013 6(x + 2) + 9<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(x + 2)4<\/sup> \u2013 6(x + 2) + 9<\/p>\n\n\n\n

(x2<\/sup> + 22<\/sup>)2<\/sup> \u2013 6x \u2013 12 + 9<\/p>\n\n\n\n

(x2<\/sup> + 22<\/sup> + 2(2)(x)) \u2013 6x \u2013 12 + 9<\/p>\n\n\n\n

x2<\/sup> + 4 + 4x \u2013 6x \u2013 12 + 9<\/p>\n\n\n\n

x2<\/sup> \u2013 2x + 1<\/p>\n\n\n\n

By using the formula (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(x \u2013 1)2<\/sup><\/p>\n\n\n\n

17. 25 \u2013 p2<\/sup> \u2013 q2<\/sup> \u2013 2pq<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

25 \u2013 p2<\/sup> \u2013 q2<\/sup> \u2013 2pq<\/p>\n\n\n\n

25 \u2013 (p2<\/sup> + q2<\/sup> + 2pq)<\/p>\n\n\n\n

(5)2<\/sup> \u2013 (p + q)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(5 + p + q) (5 \u2013p \u2013 q)<\/p>\n\n\n\n

-(p + q + 5) (p + q \u2013 5)<\/p>\n\n\n\n

18. x2<\/sup> + 9y2<\/sup> \u2013 6xy \u2013 25a2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> + 9y2<\/sup> \u2013 6xy \u2013 25a2<\/sup><\/p>\n\n\n\n

(x \u2013 3y)2<\/sup> \u2013 (5a)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(x \u2013 3y + 5a) (x \u2013 3y \u2013 5a)<\/p>\n\n\n\n

19. 49 \u2013 a2<\/sup> + 8ab \u2013 16b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

49 \u2013 a2<\/sup> + 8ab \u2013 16b2<\/sup><\/p>\n\n\n\n

49 \u2013 (a2<\/sup> \u2013 8ab + 16b2<\/sup>)<\/p>\n\n\n\n

49 \u2013 (a \u2013 4b)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a + b) (a \u2013 b)<\/p>\n\n\n\n

(7 + a \u2013 4b) (7 \u2013 a + 4b)<\/p>\n\n\n\n

-(a \u2013 4b + 7) (a \u2013 4b \u2013 7)<\/p>\n\n\n\n

20. a2<\/sup> \u2013 8ab + 16b2<\/sup> \u2013 25c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> \u2013 8ab + 16b2<\/sup> \u2013 25c2<\/sup><\/p>\n\n\n\n

(a \u2013 4b)2<\/sup>\u2013 (5c)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a \u2013 4b + 5c) (a \u2013 4b \u2013 5c)<\/p>\n\n\n\n

21. x2<\/sup> \u2013 y2<\/sup> + 6y \u2013 9<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> \u2013 y2<\/sup> + 6y \u2013 9<\/p>\n\n\n\n

x2<\/sup> + 6y \u2013 (y2<\/sup> \u2013 6y + 9)<\/p>\n\n\n\n

x2<\/sup> \u2013 (y \u2013 3)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(x + y \u2013 3) (x \u2013 y + 3)<\/p>\n\n\n\n

22. 25x2<\/sup> \u2013 10x + 1 \u2013 36y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

25x2<\/sup> \u2013 10x + 1 \u2013 36y2<\/sup><\/p>\n\n\n\n

(5x)2<\/sup> \u2013 2 (5x) + 1 \u2013 (6y)2<\/sup><\/p>\n\n\n\n

(5x \u2013 1)2<\/sup> \u2013 (6y)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(5x \u2013 6y \u2013 1) (5x + 6y \u2013 1)<\/p>\n\n\n\n

23. a2<\/sup> \u2013 b2<\/sup> + 2bc \u2013 c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> \u2013 b2<\/sup> + 2bc \u2013 c2<\/sup><\/p>\n\n\n\n

a2<\/sup> \u2013 (b2<\/sup> \u2013 2bc + c2<\/sup>)<\/p>\n\n\n\n

a2<\/sup> \u2013 (b \u2013 c)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a + b \u2013 c) (a \u2013 b + c)<\/p>\n\n\n\n

24. a2<\/sup> + 2ab + b2<\/sup> \u2013 c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 2ab + b2<\/sup> \u2013 c2<\/sup><\/p>\n\n\n\n

(a + b)2<\/sup> \u2013 c2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a + b + c) (a + b \u2013 c)<\/p>\n\n\n\n

25. 49 \u2013 x2<\/sup> \u2013 y2<\/sup> + 2xy<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

49 \u2013 x2<\/sup> \u2013 y2<\/sup> + 2xy<\/p>\n\n\n\n

49 \u2013 (x2<\/sup> + y2<\/sup> \u2013 2xy)<\/p>\n\n\n\n

72<\/sup> \u2013 (x \u2013 y)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)[7 + (x \u2013 y)] [7 \u2013 x + y]<\/p>\n\n\n\n

(x \u2013 y + 7) (y \u2013 x + 7)<\/p>\n\n\n\n

26. a2<\/sup> + 4b2<\/sup> \u2013 4ab \u2013 4c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 4b2<\/sup> \u2013 4ab \u2013 4c2<\/sup><\/p>\n\n\n\n

a2<\/sup> \u2013 2 (a) (2b) + (2b)2<\/sup> \u2013 (2c)2<\/sup><\/p>\n\n\n\n

(a \u2013 2b)2<\/sup> \u2013 (2c)2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a \u2013 2b + 2c) (a \u2013 2b \u2013 2c)<\/p>\n\n\n\n

27. x2<\/sup> \u2013 y2<\/sup> \u2013 4xz + 4z2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> \u2013 y2<\/sup> \u2013 4xz + 4z2<\/sup><\/p>\n\n\n\n

x2<\/sup> \u2013 2 (x) (2z) + (2z)2<\/sup> \u2013 y2<\/sup><\/p>\n\n\n\n

As (a-b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab<\/p>\n\n\n\n

(x \u2013 2z)2<\/sup> \u2013 y2<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(x + y \u2013 2z) (x \u2013 y \u2013 2z)<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.7 PAGE NO: 7.27<\/h3>\n\n\n\n

Factorize each of the following algebraic expressions:<\/strong><\/p>\n\n\n\n

1. x2<\/sup> + 12x \u2013 45<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> + 12x \u2013 45<\/p>\n\n\n\n

To factorize the given expression we have to find two numbers p and q such that p+q = 12 and pq = -45<\/p>\n\n\n\n

So we can replace 12x by 15x \u2013 3x<\/p>\n\n\n\n

-45 by 15 \u00d7 3<\/p>\n\n\n\n

x2<\/sup> + 12x \u2013 45 = x2<\/sup> + 15x \u2013 3x \u2013 45<\/p>\n\n\n\n

= x (x + 15) \u2013 3 (x + 15)<\/p>\n\n\n\n

= (x \u2013 3) (x + 15)<\/p>\n\n\n\n

2. 40 + 3x \u2013 x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

40 + 3x \u2013 x2<\/sup><\/p>\n\n\n\n

-(x2<\/sup> \u2013 3x \u2013 40)<\/p>\n\n\n\n

By considering, p+q = -3 and pq = -40<\/p>\n\n\n\n

So we can replace -3x by 5x \u2013 8x<\/p>\n\n\n\n

-40 by 5 \u00d7 -8<\/p>\n\n\n\n

-(x2<\/sup> \u2013 3x \u2013 40) = x2<\/sup> + 5x \u2013 8x \u2013 40<\/p>\n\n\n\n

= -x (x + 5) \u2013 8 (x + 5)<\/p>\n\n\n\n

= -(x \u2013 8) (x + 5)<\/p>\n\n\n\n

= (-x + 8) (x + 5)<\/p>\n\n\n\n

3. a2<\/sup> + 3a \u2013 88<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 3a \u2013 88<\/p>\n\n\n\n

By considering, p+q = 3 and pq = -88<\/p>\n\n\n\n

So we can replace 3a by 11a \u2013 8a<\/p>\n\n\n\n

-40 by -11 \u00d7 8<\/p>\n\n\n\n

a2<\/sup> + 3a \u2013 88 = a2<\/sup> + 11a \u2013 8a \u2013 88<\/p>\n\n\n\n

= a (a + 11) \u2013 8 (a + 11)<\/p>\n\n\n\n

= (a \u2013 8) (a + 11)<\/p>\n\n\n\n

4. a2<\/sup> \u2013 14a \u2013 51<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> \u2013 14a \u2013 51<\/p>\n\n\n\n

By considering, p+q = -14 and pq = -51<\/p>\n\n\n\n

So we can replace -14a by 3a \u2013 17a<\/p>\n\n\n\n

-51 by -17 \u00d7 3<\/p>\n\n\n\n

a2<\/sup> \u2013 14a \u2013 51 = a2<\/sup> + 3a \u2013 17a \u2013 51<\/p>\n\n\n\n

= a (a + 3) \u2013 17 (a + 3)<\/p>\n\n\n\n

= (a \u2013 17) (a + 3)<\/p>\n\n\n\n

5. x2<\/sup> + 14x + 45<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> + 14x + 45<\/p>\n\n\n\n

By considering, p+q = 14 and pq = 45<\/p>\n\n\n\n

So we can replace 14x by 5x + 9x<\/p>\n\n\n\n

45 by 5 \u00d7 9<\/p>\n\n\n\n

x2<\/sup> + 14x + 45 = x2<\/sup> + 5x + 9x + 45<\/p>\n\n\n\n

= x (x + 5) \u2013 9 (x + 5)<\/p>\n\n\n\n

= (x + 9) (x + 5)<\/p>\n\n\n\n

6. x2<\/sup> \u2013 22x + 120<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> \u2013 22x + 120<\/p>\n\n\n\n

By considering, p+q = -22 and pq = 120<\/p>\n\n\n\n

So we can replace -22x by -12x -10x<\/p>\n\n\n\n

120 by -12 \u00d7 -10<\/p>\n\n\n\n

x2<\/sup> \u2013 22x + 120 = x2<\/sup> \u2013 12x \u2013 10x + 120<\/p>\n\n\n\n

= x (x \u2013 12) \u2013 10 (x \u2013 12)<\/p>\n\n\n\n

= (x \u2013 10) (x \u2013 12)<\/p>\n\n\n\n

7. x2<\/sup> \u2013 11x \u2013 42<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> \u2013 11x \u2013 42<\/p>\n\n\n\n

By considering, p+q = -11 and pq = -42<\/p>\n\n\n\n

So we can replace -11x by 3x -14x<\/p>\n\n\n\n

-42 by 3 \u00d7 -14<\/p>\n\n\n\n

x2<\/sup> \u2013 11x \u2013 42 = x2<\/sup> + 3x \u2013 14x \u2013 42<\/p>\n\n\n\n

= x (x + 3) \u2013 14 (x + 3)<\/p>\n\n\n\n

= (x \u2013 14) (x + 3)<\/p>\n\n\n\n

8. a2<\/sup> + 2a \u2013 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 2a \u2013 3<\/p>\n\n\n\n

By considering, p+q = 2 and pq = -3<\/p>\n\n\n\n

So we can replace 2a by 3a -a<\/p>\n\n\n\n

-3 by 3 \u00d7 -1<\/p>\n\n\n\n

a2<\/sup> + 2a \u2013 3 = a2<\/sup> + 3a \u2013 a \u2013 3<\/p>\n\n\n\n

= a (a + 3) \u2013 1 (a + 3)<\/p>\n\n\n\n

= (a \u2013 1) (a + 3)<\/p>\n\n\n\n

9. a2<\/sup> + 14a + 48<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 14a + 48<\/p>\n\n\n\n

By considering, p+q = 14 and pq = 48<\/p>\n\n\n\n

So we can replace 14a by 8a + 6a<\/p>\n\n\n\n

48 by 8 \u00d7 6<\/p>\n\n\n\n

a2<\/sup> + 14a + 48 = a2<\/sup> + 8a + 6a + 48<\/p>\n\n\n\n

= a (a + 8) + 6 (a + 8)<\/p>\n\n\n\n

= (a + 6) (a + 8)<\/p>\n\n\n\n

10. x2<\/sup> \u2013 4x \u2013 21<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> \u2013 4x \u2013 21<\/p>\n\n\n\n

By considering, p+q = -4 and pq = -21<\/p>\n\n\n\n

So we can replace -4x by 3x \u2013 7x<\/p>\n\n\n\n

-21 by 3 \u00d7 -7<\/p>\n\n\n\n

x2<\/sup> + 4x \u2013 21 = x2<\/sup> + 3x \u2013 7x \u2013 21<\/p>\n\n\n\n

= x (x + 3) \u2013 7 (x + 3)<\/p>\n\n\n\n

= (x \u2013 7) (x + 3)<\/p>\n\n\n\n

11. y2<\/sup> + 5y \u2013 36<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

y2<\/sup> + 5y \u2013 36<\/p>\n\n\n\n

By considering, p+q = 5 and pq = -36<\/p>\n\n\n\n

So we can replace 5y by 9y \u2013 4y<\/p>\n\n\n\n

-36 by 9 \u00d7 -4<\/p>\n\n\n\n

y2<\/sup> + 5y \u2013 36 = y2<\/sup> + 9y \u2013 4y \u2013 36<\/p>\n\n\n\n

= y (y + 9) \u2013 4 (y + 9)<\/p>\n\n\n\n

= (y \u2013 4) (y + 9)<\/p>\n\n\n\n

12. (a2<\/sup> \u2013 5a)2<\/sup> \u2013 36<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(a2<\/sup> \u2013 5a)2<\/sup> \u2013 36<\/p>\n\n\n\n

(a2<\/sup> \u2013 5a)2<\/sup> \u2013 62<\/sup><\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

(a2<\/sup> \u2013 5a)2<\/sup> \u2013 62<\/sup> = (a2<\/sup> \u2013 5a + 6) (a2<\/sup> \u2013 5a \u2013 6)<\/p>\n\n\n\n

So now we shall factorize the expression (a2<\/sup> \u2013 5a + 6)<\/p>\n\n\n\n

By considering, p+q = -5 and pq = 6<\/p>\n\n\n\n

So we can replace -5a by a -6a<\/p>\n\n\n\n

6 by 1 \u00d7 -6<\/p>\n\n\n\n

a2<\/sup> -5a \u2013 6 = a2<\/sup> + a \u2013 6a \u2013 6<\/p>\n\n\n\n

= a (a + 1) -6(a + 1)<\/p>\n\n\n\n

= (a \u2013 6) (a + 1)<\/p>\n\n\n\n

So now we shall factorize the expression (a2<\/sup> \u2013 5a + 6)<\/p>\n\n\n\n

By considering, p+q = -5 and pq = -6<\/p>\n\n\n\n

So we can replace -5a by -2a -3a<\/p>\n\n\n\n

6 by -2 \u00d7 -3<\/p>\n\n\n\n

a2<\/sup> -5a + 6 = a2<\/sup> \u2013 2a \u2013 3a + 6<\/p>\n\n\n\n

= a (a \u2013 2) -3 (a \u2013 2)<\/p>\n\n\n\n

= (a \u2013 3) (a \u2013 2)<\/p>\n\n\n\n

\u2234 (a2<\/sup> \u2013 5a)2<\/sup> \u2013 36 = (a2<\/sup> \u2013 5a + 6) (a2<\/sup> \u2013 5a \u2013 6)<\/p>\n\n\n\n

= (a + 1) (a \u2013 6) (a \u2013 2) (a \u2013 3)<\/p>\n\n\n\n

13. (a + 7) (a \u2013 10) + 16<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(a + 7) (a \u2013 10) + 16<\/p>\n\n\n\n

a2<\/sup> \u2013 10a + 7a \u2013 70 + 16<\/p>\n\n\n\n

a2<\/sup> \u2013 3a \u2013 54<\/p>\n\n\n\n

By considering, p+q = -3 and pq = -54<\/p>\n\n\n\n

So we can replace -3a by 6a \u2013 9a<\/p>\n\n\n\n

-54 by 6 \u00d7 -9<\/p>\n\n\n\n

a2<\/sup> \u2013 3a \u2013 54 = a2<\/sup> + 6a \u2013 9a \u2013 54<\/p>\n\n\n\n

= a (a + 6) -9 (a + 6)<\/p>\n\n\n\n

= (a \u2013 9) (a + 6)<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.8 PAGE NO: 7.30<\/h4>\n\n\n\n

Resolve each of the following quadratic trinomials into factors:<\/strong><\/p>\n\n\n\n

1. 2x2<\/sup> + 5x + 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

2x2<\/sup> + 5x + 3<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 2<\/p>\n\n\n\n

The coefficient of x is 5<\/p>\n\n\n\n

Constant term is 3<\/p>\n\n\n\n

We shall split up the center term i.e., 5 into two parts such that their sum p+q is 5 and product pq = 2 \u00d7 3 is 6<\/p>\n\n\n\n

So, we express the middle term 5x as 2x + 3x<\/p>\n\n\n\n

2x2<\/sup> + 5x + 3 = 2x2<\/sup> + 2x + 3x + 3<\/p>\n\n\n\n

= 2x (x + 1) + 3 (x + 1)<\/p>\n\n\n\n

= (2x + 3) (x + 1)<\/p>\n\n\n\n

2. 2x2<\/sup> \u2013 3x \u2013 2<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

2x2<\/sup> \u2013 3x \u2013 2<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 2<\/p>\n\n\n\n

The coefficient of x is -3<\/p>\n\n\n\n

Constant term is -2<\/p>\n\n\n\n

So, we express the middle term -3x as -4x + x<\/p>\n\n\n\n

2x2<\/sup> \u2013 3x \u2013 2 = 2x2<\/sup> \u2013 4x + x \u2013 2<\/p>\n\n\n\n

= 2x (x \u2013 2) + 1 (x \u2013 2)<\/p>\n\n\n\n

= (x \u2013 2) (2x + 1)<\/p>\n\n\n\n

3. 3x2<\/sup> + 10x + 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

3x2<\/sup> + 10x + 3<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 3<\/p>\n\n\n\n

The coefficient of x is 10<\/p>\n\n\n\n

Constant term is 3<\/p>\n\n\n\n

So, we express the middle term 10x as 9x + x<\/p>\n\n\n\n

3x2<\/sup> + 10x + 3 = 3x2<\/sup> + 9x + x + 3<\/p>\n\n\n\n

= 3x (x + 3) + 1 (x + 3)<\/p>\n\n\n\n

= (3x + 1) (x + 3)<\/p>\n\n\n\n

4. 7x \u2013 6 \u2013 2x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

7x \u2013 6 \u2013 2x2<\/sup><\/p>\n\n\n\n

\u2013 2x2<\/sup> + 7x \u2013 6<\/p>\n\n\n\n

2x2<\/sup> \u2013 7x + 6<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 2<\/p>\n\n\n\n

The coefficient of x is -7<\/p>\n\n\n\n

Constant term is 6<\/p>\n\n\n\n

So, we express the middle term -7x as -4x \u2013 3x<\/p>\n\n\n\n

2x2<\/sup> \u2013 7x + 6 = 2x2<\/sup> \u2013 4x \u2013 3x + 6<\/p>\n\n\n\n

= 2x (x \u2013 2) \u2013 3 (x \u2013 2)<\/p>\n\n\n\n

= (x \u2013 2) (2x \u2013 3)<\/p>\n\n\n\n

5. 7x2<\/sup> \u2013 19x \u2013 6<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

7x2<\/sup> \u2013 19x \u2013 6<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 7<\/p>\n\n\n\n

The coefficient of x is -19<\/p>\n\n\n\n

Constant term is -6<\/p>\n\n\n\n

So, we express the middle term -19x as 2x \u2013 21x<\/p>\n\n\n\n

7x2<\/sup> \u2013 19x \u2013 6 = 7x2<\/sup> + 2x \u2013 21x \u2013 6<\/p>\n\n\n\n

= x (7x + 2) \u2013 3 (7x + 2)<\/p>\n\n\n\n

= (7x + 2) (x \u2013 3)<\/p>\n\n\n\n

6. 28 \u2013 31x \u2013 5x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

28 \u2013 31x \u2013 5x2<\/sup><\/p>\n\n\n\n

\u2013 5x2<\/sup> -31x + 28<\/p>\n\n\n\n

5x2<\/sup> + 31x \u2013 28<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 5<\/p>\n\n\n\n

The coefficient of x is 31<\/p>\n\n\n\n

Constant term is -28<\/p>\n\n\n\n

So, we express the middle term 31x as -4x + 35x<\/p>\n\n\n\n

5x2<\/sup> + 31x \u2013 28 = 5x2<\/sup> \u2013 4x + 35x \u2013 28<\/p>\n\n\n\n

= x (5x \u2013 4) + 7 (5x \u2013 4)<\/p>\n\n\n\n

= (x + 7) (5x \u2013 4)<\/p>\n\n\n\n

7. 3 + 23y \u2013 8y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

3 + 23y \u2013 8y2<\/sup><\/p>\n\n\n\n

\u2013 8y2<\/sup> + 23y + 3<\/p>\n\n\n\n

8y2<\/sup> \u2013 23y \u2013 3<\/p>\n\n\n\n

The coefficient of y2<\/sup> is 8<\/p>\n\n\n\n

The coefficient of y is -23<\/p>\n\n\n\n

Constant term is -3<\/p>\n\n\n\n

So, we express the middle term -23y as -24y + y<\/p>\n\n\n\n

8y2<\/sup> \u2013 23y \u2013 3 = 8y2<\/sup> \u2013 24y + y \u2013 3<\/p>\n\n\n\n

= 8y (y \u2013 3) + 1 (y \u2013 3)<\/p>\n\n\n\n

= (8y + 1) (y \u2013 3)<\/p>\n\n\n\n

8. 11x2<\/sup> \u2013 54x + 63<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

11x2<\/sup> \u2013 54x + 63<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 11<\/p>\n\n\n\n

The coefficient of x is -54<\/p>\n\n\n\n

Constant term is 63<\/p>\n\n\n\n

So, we express the middle term -54x as -33x \u2013 21x<\/p>\n\n\n\n

11x2<\/sup> \u2013 54x + 63 = 11x2<\/sup> \u2013 33x \u2013 21x + 63<\/p>\n\n\n\n

= 11x (x \u2013 3) \u2013 21 (x \u2013 3)<\/p>\n\n\n\n

= (11x \u2013 21) (x \u2013 3)<\/p>\n\n\n\n

9. 7x \u2013 6x2<\/sup> + 20<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

7x \u2013 6x2<\/sup> + 20<\/p>\n\n\n\n

\u2013 6x2<\/sup> + 7x + 20<\/p>\n\n\n\n

6x2<\/sup> \u2013 7x \u2013 20<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 6<\/p>\n\n\n\n

The coefficient of x is -7<\/p>\n\n\n\n

Constant term is -20<\/p>\n\n\n\n

So, we express the middle term -7x as -15x + 8x<\/p>\n\n\n\n

6x2<\/sup> \u2013 7x \u2013 20 = 6x2<\/sup> \u2013 15x + 8x \u2013 20<\/p>\n\n\n\n

= 3x (2x \u2013 5) + 4 (2x \u2013 5)<\/p>\n\n\n\n

= (3x + 4) (2x \u2013 5)<\/p>\n\n\n\n

10. 3x2<\/sup> + 22x + 35<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

3x2<\/sup> + 22x + 35<\/p>\n\n\n\n

The coefficient of x2<\/sup> is 3<\/p>\n\n\n\n

The coefficient of x is 22<\/p>\n\n\n\n

Constant term is 35<\/p>\n\n\n\n

So, we express the middle term 22x as 15x + 7x<\/p>\n\n\n\n

3x2<\/sup> + 22x + 35 = 3x2<\/sup> + 15x + 7x + 35<\/p>\n\n\n\n

= 3x (x + 5) + 7 (x + 5)<\/p>\n\n\n\n

= (3x + 7) (x+ 5)<\/p>\n\n\n\n

11. 12x2<\/sup> \u2013 17xy + 6y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

12x2<\/sup> \u2013 17xy + 6y2<\/sup><\/p>\n\n\n\n

The coefficient of x2<\/sup> is 12<\/p>\n\n\n\n

The coefficient of x is -17y<\/p>\n\n\n\n

Constant term is 6y2<\/sup><\/p>\n\n\n\n

So, we express the middle term -17xy as -9xy \u2013 8xy<\/p>\n\n\n\n

12x2<\/sup> -17xy+ 6y2<\/sup> = 12x2<\/sup> \u2013 9xy \u2013 8xy + 6y2<\/sup><\/p>\n\n\n\n

= 3x (4x \u2013 3y) \u2013 2y (4x \u2013 3y)<\/p>\n\n\n\n

= (3x \u2013 2y) (4x \u2013 3y)<\/p>\n\n\n\n

12. 6x2<\/sup> \u2013 5xy \u2013 6y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

6x2<\/sup> \u2013 5xy \u2013 6y2<\/sup><\/p>\n\n\n\n

The coefficient of x2<\/sup> is 6<\/p>\n\n\n\n

The coefficient of x is -5y<\/p>\n\n\n\n

Constant term is -6y2<\/sup><\/p>\n\n\n\n

So, we express the middle term -5xy as 4xy \u2013 9xy<\/p>\n\n\n\n

6x2<\/sup> -5xy- 6y2<\/sup> = 6x2<\/sup> + 4xy \u2013 9xy \u2013 6y2<\/sup><\/p>\n\n\n\n

= 2x (3x + 2y) -3y (3x + 2y)<\/p>\n\n\n\n

= (2x \u2013 3y) (3x + 2y)<\/p>\n\n\n\n

13. 6x2<\/sup> \u2013 13xy + 2y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

6x2<\/sup> \u2013 13xy + 2y2<\/sup><\/p>\n\n\n\n

The coefficient of x2<\/sup> is 6<\/p>\n\n\n\n

The coefficient of x is -13y<\/p>\n\n\n\n

Constant term is 2y2<\/sup><\/p>\n\n\n\n

So, we express the middle term -13xy as -12xy \u2013 xy<\/p>\n\n\n\n

6x2<\/sup> -13xy+ 2y2<\/sup> = 6x2<\/sup> \u2013 12xy \u2013 xy + 2y2<\/sup><\/p>\n\n\n\n

= 6x (x \u2013 2y) \u2013 y (x \u2013 2y)<\/p>\n\n\n\n

= (6x \u2013 y) (x \u2013 2y)<\/p>\n\n\n\n

14. 14x2<\/sup> + 11xy \u2013 15y2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

14x2<\/sup> + 11xy \u2013 15y2<\/sup><\/p>\n\n\n\n

The coefficient of x2<\/sup> is 14<\/p>\n\n\n\n

The coefficient of x is 11y<\/p>\n\n\n\n

Constant term is -15y2<\/sup><\/p>\n\n\n\n

So, we express the middle term 11xy as 21xy \u2013 10xy<\/p>\n\n\n\n

14x2<\/sup> + 11xy- 15y2<\/sup> = 14x2<\/sup> + 21xy \u2013 10xy \u2013 15y2<\/sup><\/p>\n\n\n\n

= 2x (7x \u2013 5y) + 3y (7x \u2013 5y)<\/p>\n\n\n\n

= (2x + 3y) (7x \u2013 5y)<\/p>\n\n\n\n

15. 6a2<\/sup> + 17ab \u2013 3b2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

6a2<\/sup> + 17ab \u2013 3b2<\/sup><\/p>\n\n\n\n

The coefficient of a2<\/sup> is 6<\/p>\n\n\n\n

The coefficient of a is 17b<\/p>\n\n\n\n

Constant term is -3b2<\/sup><\/p>\n\n\n\n

So, we express the middle term 17ab as 18ab \u2013 ab<\/p>\n\n\n\n

6a2<\/sup> +17ab\u2013 3b2<\/sup> = 6a2<\/sup> + 18ab \u2013 ab \u2013 3b2<\/sup><\/p>\n\n\n\n

= 6a (a + 3b) \u2013 b (a + 3b)<\/p>\n\n\n\n

= (6a \u2013 b) (a + 3b)<\/p>\n\n\n\n

16. 36a2<\/sup> + 12abc \u2013 15b2<\/sup>c2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

36a2<\/sup> + 12abc \u2013 15b2<\/sup>c2<\/sup><\/p>\n\n\n\n

The coefficient of a2<\/sup> is 36<\/p>\n\n\n\n

The coefficient of a is 12bc<\/p>\n\n\n\n

Constant term is -15b2<\/sup>c2<\/sup><\/p>\n\n\n\n

So, we express the middle term 12abc as 30abc \u2013 18abc<\/p>\n\n\n\n

36a2<\/sup> \u201312abc\u2013 15b2<\/sup>c2<\/sup> = 36a2<\/sup> + 30abc \u2013 18abc \u2013 15b2<\/sup>c2<\/sup><\/p>\n\n\n\n

= 6a (6a + 5bc) \u2013 3bc (6a + 5bc)<\/p>\n\n\n\n

= (6a + 5bc) (6a \u2013 3bc)<\/p>\n\n\n\n

= (6a + 5bc) 3(2a \u2013 bc)<\/p>\n\n\n\n

17. 15x2<\/sup> \u2013 16xyz \u2013 15y2<\/sup>z2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

15x2<\/sup> \u2013 16xyz \u2013 15y2<\/sup>z2<\/sup><\/p>\n\n\n\n

The coefficient of x2<\/sup> is 15<\/p>\n\n\n\n

The coefficient of x is -16yz<\/p>\n\n\n\n

Constant term is -15y2<\/sup>z2<\/sup><\/p>\n\n\n\n

So, we express the middle term -16xyz as -25xyz + 9xyz<\/p>\n\n\n\n

15x2<\/sup> -16xyz- 15y2<\/sup>z2<\/sup> = 15x2<\/sup> \u2013 25yz + 9yz \u2013 15y2<\/sup>z2<\/sup><\/p>\n\n\n\n

= 5x (3x \u2013 5yz) + 3yz (3x \u2013 5yz)<\/p>\n\n\n\n

= (5x + 3yz) (3x \u2013 5yz)<\/p>\n\n\n\n

18. (x \u2013 2y)2<\/sup> \u2013 5 (x \u2013 2y) + 6<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(x \u2013 2y)2<\/sup> \u2013 5 (x \u2013 2y) + 6<\/p>\n\n\n\n

The coefficient of (x-2y)2<\/sup> is 1<\/p>\n\n\n\n

The coefficient of (x-2y) is -5<\/p>\n\n\n\n

Constant term is 6<\/p>\n\n\n\n

So, we express the middle term -5(x \u2013 2y) as -2(x \u2013 2y) -3(x \u2013 2y)<\/p>\n\n\n\n

(x \u2013 2y)2<\/sup> \u2013 5 (x \u2013 2y) + 6 = (x \u2013 2y)2<\/sup> \u2013 2 (x \u2013 2y) \u2013 3 (x \u2013 2y) + 6<\/p>\n\n\n\n

= (x \u2013 2y \u2013 2) (x \u2013 2y \u2013 3)<\/p>\n\n\n\n

19. (2a \u2013 b)2<\/sup> + 2 (2a \u2013 b) \u2013 8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

(2a \u2013 b)2<\/sup> + 2 (2a \u2013 b) \u2013 8<\/p>\n\n\n\n

The coefficient of (2a-b)2<\/sup> is 1<\/p>\n\n\n\n

The coefficient of (2a-b) is 2<\/p>\n\n\n\n

Constant term is -8<\/p>\n\n\n\n

So, we express the middle term 2(2a \u2013 b) as 4 (2a \u2013b) \u2013 2 (2a \u2013 b)<\/p>\n\n\n\n

(2a \u2013 b)2<\/sup> + 2 (2a \u2013 b) \u2013 8 = (2a \u2013 b)2<\/sup> + 4 (2a \u2013 b) \u2013 2 (2a \u2013 b) \u2013 8<\/p>\n\n\n\n

= (2a \u2013 b) (2a \u2013 b + 4) \u2013 2 (2a \u2013 b + 4)<\/p>\n\n\n\n

= (2a \u2013 b + 4) (2a \u2013 b \u2013 2)<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 7.9 PAGE NO: 7.32<\/h4>\n\n\n\n

Factorize each of the following quadratic polynomials by using the method of completing the square:<\/strong><\/p>\n\n\n\n

1. p2<\/sup> + 6p + 8<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

p2<\/sup> + 6p + 8<\/p>\n\n\n\n

Coefficient of p2<\/sup> is unity. So, we add and subtract square of half of coefficient of p.<\/p>\n\n\n\n

p2<\/sup> + 6p + 8 = p2<\/sup> + 6p + 32<\/sup> \u2013 32<\/sup> + 8 (Adding and subtracting 32<\/sup>)<\/p>\n\n\n\n

= (p + 3)2<\/sup> \u2013 12<\/sup> (By completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (p + 3 \u2013 1) (p + 3 + 1)<\/p>\n\n\n\n

= (p + 2) (p + 4)<\/p>\n\n\n\n

2. q2<\/sup> \u2013 10q + 21<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

q2<\/sup> \u2013 10q + 21<\/p>\n\n\n\n

Coefficient of q2<\/sup> is unity. So, we add and subtract square of half of coefficient of q.<\/p>\n\n\n\n

q2<\/sup> \u2013 10q + 21 = q2<\/sup> \u2013 10q+ 52<\/sup> \u2013 52<\/sup> + 21 (Adding and subtracting 52<\/sup>)<\/p>\n\n\n\n

= (q \u2013 5)2<\/sup> \u2013 22<\/sup> (By completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (q \u2013 5 \u2013 2) (q \u2013 5 + 2)<\/p>\n\n\n\n

= (q \u2013 3) (q \u2013 7)<\/p>\n\n\n\n

3. 4y2<\/sup> + 12y + 5<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

4y2<\/sup> + 12y + 5<\/p>\n\n\n\n

4(y2<\/sup> + 3y + 5\/4)<\/p>\n\n\n\n

Coefficient of y2<\/sup> is unity. So, we add and subtract square of half of coefficient of y.<\/p>\n\n\n\n

4(y2<\/sup> + 3y + 5\/4) = 4 [y2<\/sup> + 3y + (3\/2)2<\/sup> \u2013 (3\/2)2<\/sup> + 5\/4] (Adding and subtracting (3\/2)2<\/sup>)<\/p>\n\n\n\n

= 4 [(y + 3\/2)2<\/sup> \u2013 12<\/sup>] (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= 4 (y + 3\/2 + 1) (y + 3\/2 \u2013 1)<\/p>\n\n\n\n

= 4 (y + 1\/2) (y + 5\/2) (by taking LCM)<\/p>\n\n\n\n

= 4 [(2y + 1)\/2] [(2y + 5)\/2]<\/p>\n\n\n\n

= (2y + 1) (2y + 5)<\/p>\n\n\n\n

4. p2<\/sup> + 6p \u2013 16<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

p2<\/sup> + 6p \u2013 16<\/p>\n\n\n\n

Coefficient of p2<\/sup> is unity. So, we add and subtract square of half of coefficient of p.<\/p>\n\n\n\n

p2<\/sup> + 6p \u2013 16 = p2<\/sup> + 6p + 32<\/sup> \u2013 32<\/sup> \u2013 16 (Adding and subtracting 32<\/sup>)<\/p>\n\n\n\n

= (p + 3)2<\/sup> \u2013 52<\/sup> (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (p + 3 + 5) (p + 3 \u2013 5)<\/p>\n\n\n\n

= (p + 8) (p \u2013 2)<\/p>\n\n\n\n

5. x2<\/sup> + 12x + 20<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

x2<\/sup> + 12x + 20<\/p>\n\n\n\n

Coefficient of x2<\/sup> is unity. So, we add and subtract square of half of coefficient of x.<\/p>\n\n\n\n

x2<\/sup> + 12x + 20 = x2<\/sup> + 12x + 62<\/sup> \u2013 62<\/sup> + 20 (Adding and subtracting 62<\/sup>)<\/p>\n\n\n\n

= (x + 6)2<\/sup> \u2013 42<\/sup> (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (x + 6 + 4) (x + 6 \u2013 4)<\/p>\n\n\n\n

= (x + 2) (x + 10)<\/p>\n\n\n\n

6. a2<\/sup> \u2013 14a \u2013 51<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> \u2013 14a \u2013 51<\/p>\n\n\n\n

Coefficient of a2<\/sup> is unity. So, we add and subtract square of half of coefficient of a.<\/p>\n\n\n\n

a2<\/sup> \u2013 14a \u2013 51 = a2<\/sup> \u2013 14a + 72<\/sup> \u2013 72<\/sup> \u2013 51 (Adding and subtracting 72<\/sup>)<\/p>\n\n\n\n

= (a \u2013 7)2<\/sup> \u2013 102<\/sup> (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (a \u2013 7 + 10) (9 \u2013 7 \u2013 10)<\/p>\n\n\n\n

= (a \u2013 17) (a + 3)<\/p>\n\n\n\n

7. a2<\/sup> + 2a \u2013 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

a2<\/sup> + 2a \u2013 3<\/p>\n\n\n\n

Coefficient of a2<\/sup> is unity. So, we add and subtract square of half of coefficient of a.<\/p>\n\n\n\n

a2<\/sup> + 2a \u2013 3 = a2<\/sup> + 2a + 12<\/sup> \u2013 12<\/sup> \u2013 3 (Adding and subtracting 12<\/sup>)<\/p>\n\n\n\n

= (a + 1)2<\/sup> \u2013 22<\/sup> (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (a + 1 + 2) (a + 1 \u2013 2)<\/p>\n\n\n\n

= (a + 3) (a \u2013 1)<\/p>\n\n\n\n

8. 4x2<\/sup> \u2013 12x + 5<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

4x2<\/sup> \u2013 12x + 5<\/p>\n\n\n\n

4(x2<\/sup> \u2013 3x + 5\/4)<\/p>\n\n\n\n

Coefficient of x2<\/sup> is unity. So, we add and subtract square of half of coefficient of x.<\/p>\n\n\n\n

4(x2<\/sup> \u2013 3x + 5\/4) = 4 [x2<\/sup> \u2013 3x + (3\/2)2<\/sup> \u2013 (3\/2)2<\/sup> + 5\/4] (Adding and subtracting (3\/2)2<\/sup>)<\/p>\n\n\n\n

= 4 [(x \u2013 3\/2)2<\/sup> \u2013 12<\/sup>] (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= 4 (x \u2013 3\/2 + 1) (x \u2013 3\/2 \u2013 1)<\/p>\n\n\n\n

= 4 (x \u2013 1\/2) (x \u2013 5\/2) (by taking LCM)<\/p>\n\n\n\n

= 4 [(2x-1)\/2] [(2x \u2013 5)\/2]<\/p>\n\n\n\n

= (2x \u2013 5) (2x \u2013 1)<\/p>\n\n\n\n

9. y2<\/sup> \u2013 7y + 12<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

y2<\/sup> \u2013 7y + 12<\/p>\n\n\n\n

Coefficient of y2<\/sup> is unity. So, we add and subtract square of half of coefficient of y.<\/p>\n\n\n\n

y2<\/sup> \u2013 7y + 12 = y2<\/sup> \u2013 7y + (7\/2)2<\/sup> \u2013 (7\/2)2<\/sup> + 12 [Adding and subtracting (7\/2)2<\/sup>]<\/p>\n\n\n\n

= (y \u2013 7\/2)2<\/sup> \u2013 (7\/2)2<\/sup> (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (y \u2013 (7\/2- 1\/2)) (y \u2013 (7\/2 + 1\/2))<\/p>\n\n\n\n

= (y \u2013 3) (y \u2013 4)<\/p>\n\n\n\n

10. z2<\/sup> \u2013 4z \u2013 12<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We have,<\/p>\n\n\n\n

z2<\/sup> \u2013 4z \u2013 12<\/p>\n\n\n\n

Coefficient of z2<\/sup> is unity. So, we add and subtract square of half of coefficient of z.<\/p>\n\n\n\n

z2<\/sup> \u2013 4z \u2013 12 = z2<\/sup> \u2013 4z + 22<\/sup> \u2013 22<\/sup> \u2013 12 [Adding and subtracting 22<\/sup>]<\/p>\n\n\n\n

= (z \u2013 2)2<\/sup> \u2013 42<\/sup> (Completing the square)<\/p>\n\n\n\n

By using the formula (a2<\/sup> \u2013 b2<\/sup>) = (a+b) (a-b)<\/p>\n\n\n\n

= (z \u2013 2 + 4) (z \u2013 2 \u2013 4)<\/p>\n\n\n\n

= (z \u2013 6) (z + 2)<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 7: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 7\u2013Factorization<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 8 Maths Chapter 7\u2013Factorization PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n