{"id":99130,"date":"2021-02-08T06:45:03","date_gmt":"2021-02-08T06:45:03","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=99130"},"modified":"2023-09-18T02:35:00","modified_gmt":"2023-09-18T02:35:00","slug":"ncert-solutions-for-9th-class-maths-chapter-2-polynomials","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-2-polynomials\/","title":{"rendered":"NCERT Solutions for 9th Class Maths :Chapter 2 Polynomials"},"content":{"rendered":"\n

Class 9: Maths Chapter 2 solutions. Complete Class 9 Maths Chapter 2 Notes.<\/p>\n\n\n\n

NCERT Solutions for 9th Class Maths :Chapter 2 Polynomials<\/strong><\/h2>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No: 32<\/p>\n\n\n\n

Exercise 2.1<\/strong><\/p>\n\n\n\n

1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.<\/strong><\/p>\n\n\n\n

(i) 4x<\/em>2<\/sup> – 3x<\/em> + 7<\/p>\n\n\n\n

(ii) y<\/em>2<\/sup> + \u221a2<\/p>\n\n\n\n

(iii) 3\u221at<\/em> + t<\/em>\u221a2<\/p>\n\n\n\n

(iv) y<\/em> + 2\/y<\/em><\/p>\n\n\n\n

(v) x<\/em>10<\/sup> + y<\/em>3<\/sup> + t<\/em>50<\/sup><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 4x<\/em>2<\/sup> – 3x<\/em> + 7
There is only one variable x<\/em> with whole number power so this polynomial in one variable.<\/p>\n\n\n\n

(ii)  y<\/em>2<\/sup> + \u221a2<\/p>\n\n\n\n

There is only one variable y<\/em> with whole number power so this polynomial in one variable.<\/p>\n\n\n\n

(iii) 3\u221a2 + t<\/em>\u221a2 <\/p>\n\n\n\n

There is only one variable t<\/em> but in 3\u221at<\/em> power of t<\/em> is 1\/2 which is not a whole number so 3\u221at<\/em> + t<\/em>\u221a2 is not a polynomial.<\/p>\n\n\n\n

(iv) y<\/em> + 2\/y<\/em><\/p>\n\n\n\n

There is only one variable y <\/em>but 2\/y<\/em> = 2y<\/em>-1 so the power is not a whole number so y<\/em> + 2\/y<\/em> is not a polynomial.<\/p>\n\n\n\n

(v) x<\/em>10<\/sup> + y<\/em>3<\/sup> + t<\/em>50<\/sup><\/p>\n\n\n\n

There are three variable x<\/em>, y<\/em> and t<\/em> and there powers are whole number so this polynomial in three variable.<\/p>\n\n\n\n

2. Write the coefficients of x<\/em>2<\/sup> in each of the following:
(i) 2 + x<\/em>2<\/sup> + x<\/em>
(ii) 2 – x<\/em>2<\/sup> + x<\/em>3<\/sup>

(iv) \u221a2x<\/em> – 1<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) coefficients of x<\/em>2<\/sup> = 1
(ii) coefficients of x<\/em>2<\/sup> = -1
(iii) coefficients of x<\/em>2<\/sup> = \u03c0\/2
(iv) coefficients of x<\/em>2<\/sup> = 0<\/p>\n\n\n\n

3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

3x<\/em>35<\/sup>+7 and 4x<\/em>100<\/sup><\/p>\n\n\n\n

4. Write the degree of each of the following polynomials:<\/strong><\/p>\n\n\n\n

(i) 5x<\/em>3<\/sup> + 4x<\/em>2<\/sup> + 7x<\/em> <\/p>\n\n\n\n

(ii) 4 \u2013 y<\/em>2<\/sup> <\/p>\n\n\n\n

(iii) 5t<\/em> \u2013 \u221a7<\/p>\n\n\n\n

(iv) 3<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 5x<\/em>3<\/sup> has highest power in the given polynomial which power is 3. Therefore, degree of polynomial is 3.<\/p>\n\n\n\n

(ii) \u2013 y<\/em>2<\/sup>  has highest power in the given polynomial which power is 2. Therefore, degree of polynomial is 2.<\/p>\n\n\n\n

(iii) 5t<\/em> has highest power in the given polynomial which power is 1. Therefore, degree of polynomial is 1.<\/p>\n\n\n\n

(iv) There is no variable in the given polynomial. Therefore, degree of polynomial is 0.<\/p>\n\n\n\n

5. Classify the following as linear, quadratic and cubic polynomial:<\/strong><\/p>\n\n\n\n

(i) x<\/em>2<\/sup> + x<\/em>
\u25ba Quadratic Polynomial<\/p>\n\n\n\n

(ii) x<\/em> – x<\/em>3<\/sup>
\u25ba Cubic Polynomial<\/p>\n\n\n\n

(iii) y<\/em> + y<\/em>2<\/sup> +4
\u25ba Quadratic Polynomial<\/p>\n\n\n\n

(iv) 1 + x<\/em>
\u25ba Linear Polynomial<\/p>\n\n\n\n

(vi) r<\/em>2<\/sup><\/p>\n\n\n\n

\u25ba Quadratic Polynomial<\/p>\n\n\n\n

(vii) 7x<\/em>3<\/sup>
\u25ba Cubic Polynomial<\/p>\n\n\n\n

Page No: 34<\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

Exercise 2.2<\/strong><\/p>\n\n\n\n

1. Find the value of the polynomial at 5x <\/em>+ 4x<\/em>2<\/sup> + 3 at<\/strong>

(i) x<\/em> = 0 (ii) x<\/em> = – 1 (iii) x<\/em> = 2<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) p<\/em>(x<\/em>) = 5x <\/em>+ 4x<\/em>2<\/sup> + 3<\/p>\n\n\n\n

p<\/em>(0) = 5(0) + 4(0)2<\/sup> + 3
           = 3<\/p>\n\n\n\n

(ii) p<\/em>(x<\/em>) = 5x <\/em>+ 4x<\/em>2<\/sup> + 3<\/p>\n\n\n\n

p<\/em>(-1) = 5(-1) + 4(-1)2<\/sup> + 3
           = -5 + 4(1) + 3 = 2<\/p>\n\n\n\n

(iii) p<\/em>(x<\/em>) = 5x <\/em>+ 4x<\/em>2<\/sup> + 3<\/p>\n\n\n\n

p<\/em>(2) = 5(2) + 4(2)2<\/sup> + 3
           = 10 + 16 + 3 = 29<\/p>\n\n\n\n

2. Find p<\/em>(0), p<\/em>(1) and p<\/em>(2) for each of the following polynomials:<\/strong><\/p>\n\n\n\n

(i) p<\/em>(y<\/em>) = y<\/em>2<\/sup> – y<\/em> + 1<\/p>\n\n\n\n

(ii) p<\/em>(t<\/em>) = 2 + t<\/em> + 2t<\/em>2<\/sup> – t<\/em>3<\/sup>
(iii) p<\/em>(x<\/em>) = x<\/em>3<\/sup> <\/p>\n\n\n\n

(iv) p<\/em>(x<\/em>) = (x<\/em> – 1) (x<\/em> + 1)<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) p<\/em>(y<\/em>) = y<\/em>2<\/sup> – y<\/em> + 1
p<\/em>(0) = (0)2<\/sup> – (0) + 1 = 1
p<\/em>(1) = (1)2<\/sup> – (1) + 1 = 1
p<\/em>(2) = (2)2<\/sup> – (2) + 1 = 3<\/p>\n\n\n\n

(ii) p<\/em>(t<\/em>) = 2 + t<\/em> + 2t<\/em>2<\/sup> – t<\/em>3<\/sup>
p<\/em>(0) = 2 + 0 + 2 (0)2<\/sup> – (0)3<\/sup> = 2
p<\/em>(1) = 2 + (1) + 2(1)2<\/sup> – (1)3
= 2 + 1 + 2 – 1 = 4
p<\/em>(2) = 2 + 2 + 2(2)2<\/sup> – (2)3<\/sup>
= 2 + 2 + 8 – 8 = 4<\/p>\n\n\n\n

(iii) p<\/em>(x<\/em>) = x<\/em>3<\/sup>
p<\/em>(0) = (0)3<\/sup> = 0
p<\/em>(1) = (1)3<\/sup> = 1
p<\/em>(2) = (2)3<\/sup> = 8<\/p>\n\n\n\n

(iv) p<\/em>(x<\/em>) = (x<\/em> – 1) (x<\/em> + 1)
p<\/em>(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1
p<\/em>(1) = (1 – 1) (1 + 1) = 0 (2) = 0
p<\/em>(2) = (2 – 1 ) (2 + 1) = 1(3) = 3<\/p>\n\n\n\n

Page No: 35<\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

3. Verify whether the following are zeroes of the polynomial, indicated against them.<\/strong><\/p>\n\n\n\n

(i) p<\/em>(x<\/em>) = 3x<\/em> + 1, x<\/em> = -1\/3<\/p>\n\n\n\n

(ii)  p<\/em>(x<\/em>) = 5x<\/em> – \u03c0, x<\/em> = 4\/5<\/p>\n\n\n\n

(iii) p<\/em>(x<\/em>) = x<\/em>2<\/sup> – 1, x<\/em> = 1, -1<\/p>\n\n\n\n

(iv) p(x<\/em>) = (x<\/em> + 1) (x<\/em> – 2), x<\/em> = -1, 2
(v) p(x<\/em>) = x<\/em>2<\/sup> , x<\/em> = 0

(viii) p<\/em>(x<\/em>) = 2x<\/em> + 1, x<\/em> = 1\/2<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) If x<\/em> = -1\/3 is a zero of polynomial p<\/em>(x<\/em>) = 3x<\/em> + 1 then p<\/em>(-1\/3) should be 0.
At, p<\/em>(-1\/3) = 3(-1\/3) + 1 = -1 + 1 = 0
Therefore, x<\/em> = -1\/3 is a zero of polynomial p<\/em>(x<\/em>) = 3x<\/em> + 1.<\/p>\n\n\n\n

(ii) If x<\/em> = 4\/5 is a zero of polynomial p<\/em>(x<\/em>) = 5x<\/em> – \u03c0 then p<\/em>(4\/5) should be 0.
At, p<\/em>(4\/5) = 5(4\/5) – \u03c0 = 4 – \u03c0
Therefore, x<\/em> = 4\/5 is not a zero of given polynomial p<\/em>(x<\/em>) = 5x<\/em> – \u03c0.<\/p>\n\n\n\n

(iii) If x<\/em> = 1 and x<\/em> = -1 are zeroes of polynomial p<\/em>(x<\/em>) = x<\/em>2<\/sup> – 1, then p<\/em>(1) and p<\/em>(-1) should be 0.
At, p<\/em>(1) = (1)2<\/sup> – 1 = 0 and
At, p<\/em>(-1) = (-1)2<\/sup> – 1 = 0
Hence, x<\/em> = 1 and -1 are zeroes of the polynomial  p<\/em>(x<\/em>) = x<\/em>2<\/sup> – 1.<\/p>\n\n\n\n

(iv) If x<\/em> = -1 and x<\/em> = 2 are zeroes of polynomial p<\/em>(x<\/em>) = (x<\/em> +1) (x<\/em> – 2), then p<\/em>( – 1) and (2)should be 0.
At, p<\/em>(-1) = (-1 + 1) (-1 – 2) = 0 (-3) = 0, and
At, p<\/em>(2) = (2 + 1) (2 – 2) = 3 (0) = 0
Therefore, x<\/em> = -1 and x<\/em> = 2 are zeroes of the polynomial p<\/em>(x<\/em>) = (x<\/em> +1) (x<\/em> – 2).<\/p>\n\n\n\n

(v) If x<\/em> = 0 is a zero of polynomial p<\/em>(x<\/em>) = x<\/em>2<\/sup>, then p<\/em>(0) should be zero.
Here, p<\/em>(0) = (0)2<\/sup> = 0
Hence, x<\/em> = 0 is a zero of the polynomial p<\/em>(x<\/em>) = x<\/em>2<\/sup>.<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

(viii)  If x<\/em> = 1\/2 is a zero of polynomial p<\/em>(x<\/em>) = 2x<\/em> + 1 then p<\/em>(1\/2) should be 0.
At, p<\/em>(1\/2) = 2(1\/2) + 1 = 1 + 1 = 2
Therefore, x<\/em> = 1\/2 is not a zero of given polynomial p<\/em>(x<\/em>) = 2x<\/em> + 1.<\/p>\n\n\n\n

4. Find the zero of the polynomial in each of the following cases:
(i) p<\/em>(x<\/em>) = x<\/em> + 5 <\/p>\n\n\n\n

(ii) p<\/em>(x<\/em>) = x<\/em> – 5 <\/p>\n\n\n\n

(iii) p<\/em>(x<\/em>) = 2x<\/em> + 5
(iv) p<\/em>(x<\/em>) = 3x<\/em> – 2 <\/p>\n\n\n\n

(v) p<\/em>(x<\/em>) = 3x<\/em> <\/p>\n\n\n\n

(vi) p<\/em>(x<\/em>) = ax<\/em>, a<\/em> \u2260 0
(vii) p<\/em>(x<\/em>) = cx<\/em> + d<\/em>, c<\/em> \u2260 0, c<\/em>, are real numbers.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) p<\/em>(x<\/em>) = x<\/em> + 5 <\/p>\n\n\n\n

p<\/em>(x<\/em>) = 0<\/p>\n\n\n\n

x<\/em> + 5 = 0<\/p>\n\n\n\n

x<\/em> = -5<\/p>\n\n\n\n

Therefore, x<\/em> = -5 is a zero of polynomial p<\/em>(x<\/em>) = x<\/em> + 5 .<\/p>\n\n\n\n

(ii) p<\/em>(x<\/em>) = x<\/em> – 5<\/p>\n\n\n\n

p<\/em>(x<\/em>) = 0<\/p>\n\n\n\n

x<\/em> – 5 = 0<\/p>\n\n\n\n

x<\/em> = 5<\/p>\n\n\n\n

Therefore, x<\/em> = 5 is a zero of polynomial p<\/em>(x<\/em>) = x<\/em> – 5.<\/p>\n\n\n\n

(iii) p<\/em>(x<\/em>) = 2x<\/em> + 5<\/p>\n\n\n\n

p<\/em>(x<\/em>) = 0<\/p>\n\n\n\n

2x<\/em> + 5 = 0
2x<\/em> = -5
x<\/em> = -5\/2
Therefore, x<\/em> = -5\/2 is a zero of polynomial p<\/em>(x<\/em>) = 2x<\/em> + 5.<\/p>\n\n\n\n

(iv) p<\/em>(x<\/em>) = 3x<\/em> – 2
p<\/em>(x<\/em>) = 0
3x<\/em> – 2 = 0
x<\/em> = 2\/3
Therefore, x<\/em> = 2\/3 is a zero of polynomial p<\/em>(x<\/em>) = 3x<\/em> – 2.<\/p>\n\n\n\n

(v) p<\/em>(x<\/em>) = 3x<\/em>
p<\/em>(x<\/em>) = 0
3x<\/em> = 0
x<\/em> = 0
Therefore, x<\/em> = 0 is a zero of polynomial p<\/em>(x<\/em>) = 3x<\/em>.<\/p>\n\n\n\n

(vi) p<\/em>(x<\/em>) = ax<\/em>
p<\/em>(x<\/em>) = 0
ax <\/em>= 0
x <\/em>= 0
Therefore, x<\/em> = 0 is a zero of polynomial p<\/em>(x<\/em>) = ax<\/em>.<\/p>\n\n\n\n

(vii) p<\/em>(x<\/em>) = cx<\/em> + d<\/em>
p<\/em>(x<\/em>) = 0
cx<\/em> + d<\/em> = 0
x<\/em> = –d<\/em>\/c
Therefore, x<\/em> = –d<\/em>\/c is a zero of polynomial p<\/em>(x<\/em>) = cx<\/em> + d<\/em>.<\/p>\n\n\n\n

Page No: 40<\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

Exercises 2.3<\/strong><\/p>\n\n\n\n

1. Find the remainder when x<\/em>3<\/sup> + 3x<\/em>2<\/sup> + 3x<\/em> + 1 is divided by<\/strong>
(i) x<\/em> + 1
(ii) x<\/em> – 1\/2
(iii) x<\/em>
(iv) x<\/em> + \u03c0
(v) 5 + 2x<\/em><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) x<\/em> + 1
By long division,

Therefore, the remainder is 0.<\/p>\n\n\n\n

(ii) x<\/em> – 1\/2
By long division,
<\/p>\n\n\n\n

Therefore, the remainder is 27\/8.<\/p>\n\n\n\n

(iii) x<\/em>

Therefore, the remainder is 1.<\/p>\n\n\n\n

(iv) x<\/em> + \u03c0

Therefore, the remainder is [1 – 3\u03c0 + 3\u03c02<\/sup> – \u03c03<\/sup>].<\/p>\n\n\n\n

(v) 5 + 2x<\/em>

Therefore, the remainder is -27\/8.<\/p>\n\n\n\n

2. Find the remainder when x<\/em>3<\/sup> – ax<\/em>2<\/sup> + 6x<\/em> – a<\/em> is divided by x<\/em> – a<\/em>.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

By Long Division,

Therefore, remainder obtained is 5a <\/em>when x<\/em>3<\/sup> – ax<\/em>2<\/sup> + 6x<\/em> – a<\/em> is divided by x<\/em> – a<\/em>.<\/p>\n\n\n\n

3. Check whether 7 + 3x<\/em> is a factor of 3x<\/em>3<\/sup> + 7x<\/em>.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

We have to divide 3x<\/em>3<\/sup> + 7x <\/em>by 7 + 3x<\/em>. If remainder comes out to be 0 then 7 + 3x<\/em> will be a factor of 3x<\/em>3<\/sup> + 7x<\/em>.
By Long Division,

As remainder is not zero so 7 + 3x<\/em> is not a factor of 3x<\/em>3<\/sup> + 7x<\/em>.<\/p>\n\n\n\n

Page No: 43<\/p>\n\n\n\n

NCERT 10th Maths Chapter 2, class 10 Maths Chapter 2 solutions<\/p>\n\n\n\n

Exercise 2.4<\/strong><\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

1. Determine which of the following polynomials has (x<\/em> + 1) a factor:<\/strong>
(i) x<\/em>3<\/sup> + x<\/em>2<\/sup> + x<\/em> + 1<\/p>\n\n\n\n

(ii) x<\/em>4<\/sup> + x<\/em>3<\/sup> + x<\/em>2<\/sup> + x<\/em> + 1
(iii) x<\/em>4<\/sup> + 3x<\/em>3<\/sup> + 3x<\/em>2<\/sup> + x<\/em> + 1 <\/p>\n\n\n\n

(iv) x<\/em>3<\/sup> – x<\/em>2<\/sup> – (2 + \u221a2)x<\/em> + \u221a2<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) If (x<\/em> + 1) is a factor of p<\/em>(x<\/em>) = x<\/em>3<\/sup> + x<\/em>2<\/sup> + x<\/em> + 1, p<\/em>(-1) must be zero. 
Here, p<\/em>(x<\/em>) = x<\/em>3<\/sup> + x<\/em>2<\/sup> + x<\/em> + 1 
p<\/em>(-1) = (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 1 
= -1 + 1 – 1 + 1 = 0
Therefore, x<\/em> + 1 is a factor of this polynomial<\/p>\n\n\n\n

(ii) If (x<\/em> + 1) is a factor of p<\/em>(x<\/em>) = x<\/em>4<\/sup> + x<\/em>3<\/sup> + x<\/em>2<\/sup> + x<\/em> + 1, p<\/em>(-1) must be zero. 
Here, p<\/em>(x<\/em>) = x<\/em>4<\/sup> + x<\/em>3<\/sup> + x<\/em>2<\/sup> + x<\/em> + 1 
p<\/em>(-1) = (-1)4<\/sup> + (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 1
= 1 – 1 + 1 – 1 + 1 = 1<\/p>\n\n\n\n

As, p<\/em>(-1) \u2260 0
Therefore, x<\/em> + 1 is not a factor of this polynomial<\/p>\n\n\n\n

(iii)If (x<\/em> + 1) is a factor of polynomial p<\/em>(x<\/em>) = x<\/em>4<\/sup> + 3x<\/em>3<\/sup> + 3x<\/em>2<\/sup> + x<\/em> + 1, p<\/em>(- 1) must be 0. 
p<\/em>(-1) = (-1)4<\/sup> + 3(-1)3<\/sup> + 3(-1)2<\/sup> + (-1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
As, p<\/em>(-1) \u2260 0
Therefore, x<\/em> + 1 is not a factor of this polynomial.<\/p>\n\n\n\n

(iv) If (x<\/em> + 1) is a factor of polynomial<\/p>\n\n\n\n

p<\/em>(x<\/em>) = x<\/em>3<\/sup> – x<\/em>2<\/sup> – (2 + \u221a2)x<\/em> + \u221a2, p<\/em>(- 1) must be 0.<\/p>\n\n\n\n

p<\/em>(-1) =  (-1)3<\/sup> –  (-1)2<\/sup> –  (2 + \u221a2) (-1) + \u221a2
= -1 – 1 + 2 + \u221a2 + \u221a2
=2\u221a2
As, p<\/em>(-1) \u2260 0
Therefore,, x<\/em> + 1 is not a factor of this polynomial.<\/p>\n\n\n\n

2. Use the Factor Theorem to determine whether g<\/em>(x<\/em>) is a factor of p<\/em>(x<\/em>) in each of the following cases:<\/strong>
(i) p<\/em>(x<\/em>) = 2x<\/em>3<\/sup> + x<\/em>2<\/sup> – 2x<\/em> – 1, g<\/em>(x<\/em>) = x<\/em> + 1
(ii) p<\/em>(x<\/em>) = x<\/em>3<\/sup> + 3x<\/em>2<\/sup> + 3x<\/em> + 1, g<\/em>(x<\/em>) = x<\/em> + 2
(iii) p(x<\/em>) = x<\/em>3<\/sup> – 4 x<\/em>2<\/sup> + x<\/em> + 6, g<\/em>(x<\/em>) = x<\/em> – 3<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) If g<\/em>(x<\/em>) = x<\/em> + 1 is a factor of given polynomial p<\/em>(x<\/em>), p<\/em>(- 1) must be zero.
p<\/em>(x<\/em>) = 2x<\/em>3<\/sup> + x<\/em>2<\/sup> – 2x<\/em> – 1
p<\/em>(- 1) = 2(-1)3<\/sup> + (-1)2<\/sup> – 2(-1) – 1
= 2(- 1) + 1 + 2 – 1 = 0
Hence, g<\/em>(x<\/em>) = x<\/em> + 1 is a factor of given polynomial.<\/p>\n\n\n\n

(ii) If g<\/em>(x<\/em>) = x<\/em> + 2 is a factor of given polynomial p<\/em>(x<\/em>), p<\/em>(- 2) must be 0.
p<\/em>(x<\/em>) = x<\/em>3<\/sup> +3x<\/em>2<\/sup> + 3x<\/em> + 1
p<\/em>(-2) = (-2)3<\/sup> + 3(- 2)2<\/sup> + 3(- 2) + 1
= -8 + 12 – 6 + 1
= -1<\/p>\n\n\n\n

As, p<\/em>(-2) \u2260 0<\/p>\n\n\n\n

Hence g<\/em>(x<\/em>) = x<\/em> + 2 is not a factor of given polynomial.<\/p>\n\n\n\n

(iii) If g<\/em>(x<\/em>) = x<\/em> – 3 is a factor of given polynomial p<\/em>(x<\/em>), p<\/em>(3) must be 0.
p<\/em>(x<\/em>) = x<\/em>3<\/sup> – 4x<\/em>2<\/sup> + x<\/em> + 6
p<\/em>(3) = (3)3<\/sup> – 4(3)2<\/sup> + 3 + 6
= 27 – 36 + 9 = 0
Therefore,, g<\/em>(x<\/em>) = x<\/em> – 3 is a factor of given polynomial.<\/p>\n\n\n\n

Page No: 44<\/p>\n\n\n\n

3. Find the value of k<\/em>, if x<\/em> – 1 is a factor of p<\/em>(x<\/em>) in each of the following cases:<\/strong>(i) p<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + k<\/em>
(ii) p<\/em>(x<\/em>) = 2x<\/em>2<\/sup> + kx<\/em> +  \u221a2
(iii) p<\/em>(x<\/em>) = kx<\/em>2<\/sup> – \u221a2x<\/em> + 1
(iv) p<\/em>(x<\/em>) = kx<\/em>2<\/sup> – 3x<\/em> + k<\/em><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) If x<\/em> – 1 is a factor of polynomial p<\/em>(x<\/em>) = x<\/em>2<\/sup> + x<\/em> + k<\/em>, then<\/p>\n\n\n\n

p<\/em>(1) = 0
\u21d2 (1)2<\/sup> + 1 + k<\/em> = 0
\u21d2 2 + k<\/em> = 0
\u21d2 k<\/em> = – 2<\/p>\n\n\n\n

Therefore, value of k<\/em> is -2.<\/p>\n\n\n\n

(ii) If x<\/em> – 1 is a factor of polynomial p<\/em>(x<\/em>) = 2x<\/em>2<\/sup> + kx<\/em> +  \u221a2, then
p<\/em>(1) = 0
\u21d2 2(1)2<\/sup> + k<\/em>(1) + \u221a2 = 0
\u21d2 2 + k + <\/em>\u221a2 = 0
\u21d2 k<\/em> = -2 – \u221a2 = -(2 + \u221a2)<\/p>\n\n\n\n

Therefore, value of k<\/em> is -(2 + \u221a2).<\/p>\n\n\n\n

(iii) If x<\/em> – 1 is a factor of polynomial p<\/em>(x<\/em>) = kx<\/em>2<\/sup> – \u221a2x<\/em> + 1, then
p<\/em>(1) = 0
\u21d2 k<\/em>(1)2<\/sup> – \u221a2(1) + 1 = 0
\u21d2 k<\/em> – \u221a2 + 1 = 0
\u21d2 k<\/em> = \u221a2 – 1<\/p>\n\n\n\n

Therefore, value of k<\/em> is \u221a2 – 1.<\/p>\n\n\n\n

(iv) If x<\/em> – 1 is a factor of polynomial p<\/em>(x<\/em>) = kx<\/em>2<\/sup> – 3x<\/em> + k<\/em>, then
p<\/em>(1) = 0
\u21d2 k<\/em>(1)2<\/sup> + 3(1) + k<\/em> = 0
\u21d2 k<\/em> – 3 + k<\/em> = 0
\u21d2 2k<\/em> – 3 = 0
\u21d2 k<\/em> = 3\/2<\/p>\n\n\n\n

Therefore, value of k<\/em> is 3\/2.<\/p>\n\n\n\n

4. Factorise:
(i) 12x<\/em>2<\/sup> + 7x<\/em> + 1
(ii) 2x<\/em>2<\/sup> + 7x<\/em> + 3
(iii) 6x<\/em>2<\/sup> + 5x<\/em> – 6
(iv) 3x<\/em>2<\/sup> – x<\/em> – 4 <\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 12x<\/em>2<\/sup> + 7x<\/em> + 1
= 12x<\/em>2<\/sup> – 4x<\/em> – 3x<\/em>+ 1                   
= 4x<\/em> (3x<\/em> – 1) – 1 (3x<\/em> – 1)
= (3x<\/em> – 1) (4x<\/em> – 1)<\/p>\n\n\n\n

(ii) 2x<\/em>2<\/sup> + 7x<\/em> + 3
= 2x<\/em>2<\/sup> + 6x<\/em> + x <\/em>+ 3
= 2x<\/em> (x<\/em> + 3) + 1 (x<\/em> + 3)
=  (x<\/em> + 3) (2x <\/em>+ 1) <\/p>\n\n\n\n

(iii) 6x<\/em>2<\/sup> + 5x<\/em> – 6
= 6x<\/em>2<\/sup> + 9x <\/em>– 4x<\/em> – 6<\/p>\n\n\n\n

 = 3x<\/em> (2x<\/em> + 3) – 2 (2x<\/em> + 3)
= (2x<\/em> + 3) (3x<\/em> – 2)<\/p>\n\n\n\n

(iv) 3x<\/em>2<\/sup> – x<\/em> – 4
= 3x<\/em>2<\/sup> – 4x <\/em>+ 3x<\/em> – 4 
= x<\/em> (3x<\/em> – 4) + 1 (3x<\/em> – 4)
= (3x<\/em> – 4) (x<\/em> + 1)<\/p>\n\n\n\n

5. Factorise:
(i) x<\/em>3<\/sup> – 2x<\/em>2<\/sup> – x<\/em> + 2<\/p>\n\n\n\n

(ii) x<\/em>3<\/sup> – 3x<\/em>2<\/sup> – 9x<\/em> – 5
(iii) x<\/em>3<\/sup> + 13x<\/em>2<\/sup> + 32x<\/em> + 20 <\/p>\n\n\n\n

(iv) 2y<\/em>3<\/sup> + y<\/em>2<\/sup> – 2y<\/em> – 1<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Let p<\/em>(x<\/em>) = x<\/em>3<\/sup> – 2x<\/em>2<\/sup> – x<\/em> + 2
Factors of 2 are \u00b11 and \u00b1 2
By trial method, we find that
p(1)<\/em> = 0
So, (x+1)<\/em> is factor of p<\/em>(x<\/em>)
Now,
p<\/em>(x<\/em>) = x<\/em>3<\/sup> – 2x<\/em>2<\/sup> – x<\/em> + 2
p<\/em>(-1<\/em>) = (-1)3<\/sup> – 2(-1)2<\/sup> –<\/em> (-1) + 2 = -1 -2 + 1 + 2 = 0
Therefore, (x<\/em>+1) is the factor of  p<\/em>(x<\/em>)<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Now, Dividend = Divisor \u00d7 Quotient + Remainder<\/p>\n\n\n\n

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

= (x+1) (x2<\/sup> – x – 2x + 2)<\/em><\/p>\n\n\n\n

= (x+1) {x(x-1) -2(x-1)}<\/em><\/p>\n\n\n\n

= (x+1) (x-1) (x+2)<\/em><\/p>\n\n\n\n

(ii) Let p<\/em>(x<\/em>) = x<\/em>3<\/sup> – 3x<\/em>2<\/sup> – 9x<\/em> – 5
Factors of 5 are \u00b11 and \u00b15
By trial method, we find that
p(5)<\/em> = 0
So, (x-5)<\/em> is factor of p<\/em>(x<\/em>)
Now,
p<\/em>(x<\/em>) = x<\/em>3<\/sup> – 2x<\/em>2<\/sup> – x<\/em> + 2
p<\/em>(5<\/em>) = (5)3<\/sup> – 3(5)2<\/sup> –<\/em> 9(5) – 5 = 125 – 75 – 45 – 5 = 0
Therefore, (x<\/em>-5) is the factor of  p<\/em>(x<\/em>)<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Now, Dividend = Divisor \u00d7 Quotient + Remainder<\/p>\n\n\n\n

(x-5) (x2<\/sup> + 2x + 1)<\/em><\/p>\n\n\n\n

= (x-5) (<\/em>x2<\/sup> + x + x + 1)<\/em><\/p>\n\n\n\n

= <\/em>(x-5) {x(x+1) +1(x+1)}<\/em><\/p>\n\n\n\n

= <\/em>(x-5) <\/em>(x+1) <\/em>(x+1)<\/em><\/p>\n\n\n\n

(iii) Let p<\/em>(x<\/em>) = x<\/em>3<\/sup> + 13x<\/em>2<\/sup> + 32x<\/em> + 20
Factors of 20 are \u00b11, \u00b12, \u00b14, \u00b15, \u00b110 and \u00b120
By trial method, we find that
p(-1)<\/em> = 0
So, (x+1)<\/em> is factor of p<\/em>(x<\/em>)
Now,
p<\/em>(x<\/em>) =  x<\/em>3<\/sup> + 13x<\/em>2<\/sup> + 32x<\/em> + 20
p<\/em>(-1<\/em>) = (-1)3<\/sup> + 13(-1)2<\/sup> + 32<\/em>(-1) + 20 = -1 + 13 – 32 + 20 = 0
Therefore, (x<\/em>+1) is the factor of  p<\/em>(x<\/em>)<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Now, Dividend = Divisor \u00d7 Quotient + Remainder<\/p>\n\n\n\n

(x+1) (x2<\/sup> + 12x + 20)<\/em><\/p>\n\n\n\n

= (x+1) (<\/em>x2<\/sup> + 2x + 10x + 20)<\/em><\/p>\n\n\n\n

= <\/em>(x-5) {x(x+2) +10(x+2)}<\/em><\/p>\n\n\n\n

= <\/em>(x-5) <\/em>(x+2) <\/em>(x+10)<\/em><\/p>\n\n\n\n

(iv) Let p<\/em>(y<\/em>) = 2y<\/em>3<\/sup> + y<\/em>2<\/sup> – 2y<\/em> – 1
Factors of ab = 2\u00d7 (-1) = -2 are \u00b11 and \u00b12
By trial method, we find that
p(1)<\/em> = 0
So, (y-1)<\/em> is factor of p<\/em>(y<\/em>)
Now,
p<\/em>(y<\/em>) =  2y<\/em>3<\/sup> + y<\/em>2<\/sup> – 2y<\/em> – 1
p<\/em>(1<\/em>) = 2(1)3<\/sup> + (1)2<\/sup> – 2<\/em>(1) – 1 = 2 +1 – 2 – 1 = 0
Therefore, (y<\/em>-1) is the factor of  p<\/em>(y<\/em>)<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

 Now, Dividend = Divisor \u00d7 Quotient + Remainder<\/p>\n\n\n\n

(y-1) (2y2<\/sup> + 3y + 1)<\/em><\/p>\n\n\n\n

= <\/em>(y-1) (<\/em>2y2<\/sup> + 2y + y + 1)<\/em><\/p>\n\n\n\n

= <\/em>(y-1) {2y(y+1) +1(y+1)}<\/em><\/p>\n\n\n\n

= <\/em>(y-1) <\/em>(2y+1) <\/em>(y+1)<\/em><\/p>\n\n\n\n

Page No: 48<\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

Exercise 2.5<\/strong><\/p>\n\n\n\n

1. Use suitable identities to find the following products:<\/strong>
    (i) (x<\/em> + 4) (x<\/em> + 10)                     (ii) (x<\/em> + 8) (x<\/em> \u2013 10)                      (iii) (3x<\/em> + 4) (3x<\/em> \u2013 5)
    (iv) (y2 <\/sup><\/em>+ 3\/2) (y2 <\/sup><\/em>– 3\/2)             (v) (3 – 2x<\/em>) (3 + 2x<\/em>)<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Using identity, (x <\/em>+ a) (x<\/em> + b) = x<\/em>2<\/sup> + (a + b) x<\/em> + ab 
In (x<\/em> + 4) (x<\/em> + 10), a = 4 and b = 10
Now,
(x<\/em> + 4) (x<\/em> + 10) = x<\/em>2<\/sup> + (4 + 10)x<\/em> + (4 \u00d7 10)
                         = x<\/em>2<\/sup> + 14x<\/em>+ 40
(ii) (x<\/em> + 8) (x<\/em> \u2013 10)
Using identity, (x <\/em>+ a) (x<\/em> + b) = x<\/em>2<\/sup> + (a + b) x<\/em> + ab
Here, a = 8 and b = \u201310
(x<\/em> + 8) (x<\/em> \u2013 10) = x<\/em>2<\/sup> + {8 +(\u2013 10)}x<\/em> + {8\u00d7(\u2013 10)}
                         = x<\/em>2<\/sup> + (8 \u2013 10)x<\/em> \u2013 80
                         = x<\/em>2<\/sup> \u2013 2x<\/em> \u2013 80<\/p>\n\n\n\n

(iii) (3x<\/em> + 4) (3x<\/em> \u2013 5)
Using identity, (x <\/em>+ a) (x<\/em> + b) = x<\/em>2<\/sup> + (a + b) x<\/em> + ab
Here, x<\/em> = 3x , <\/em>a = 4 and b = -5
(3x<\/em> + 4) (3x<\/em> \u2013 5) = (3x<\/em>) 2<\/sup> + {4 + (-5)}3x<\/em> + {4\u00d7(-5)}
                           = 9x<\/em>2<\/sup> + 3x<\/em>(4 – 5) – 20
                           = 9x<\/em>2<\/sup> – 3x<\/em> – 20<\/p>\n\n\n\n

(iv) (y2 <\/sup><\/em>+ 3\/2) (y2 <\/sup><\/em>– 3\/2)
Using identity, (x <\/em>+ y<\/em>) (x<\/em> –y<\/em>) = x<\/em>2<\/sup> – y<\/em>2<\/sup>
Here, x<\/em> =y2<\/sup> <\/em>and y <\/em>= 3\/2
(y2 <\/sup><\/em>+ 3\/2) (y2 <\/sup><\/em>– 3\/2) = (y2<\/sup><\/em>)2 <\/sup><\/em>– (3\/2)2<\/sup><\/em>
= y<\/em>4<\/sup>– 9\/4<\/p>\n\n\n\n

(v) (3 – 2x<\/em>) (3 + 2x<\/em>)
Using identity, (x <\/em>+ y<\/em>) (x<\/em> –y<\/em>) = x<\/em>2<\/sup> – y<\/em>2<\/sup>
Here, x<\/em> = 3 and y <\/em>= 2x<\/em>
(3 – 2x<\/em>) (3 + 2x<\/em>) = 32<\/sup> – (2x<\/em>)2<\/sup>
=9- 4x<\/em>2<\/sup><\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

2. Evaluate the following products without multiplying directly:<\/strong>
    (i) 103 \u00d7 107               (ii) 95 \u00d7 96               (iii) 104 \u00d7 96<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 103 \u00d7 107 = (100 + 3) (100 + 7)
Using identity, (x <\/em>+ a) (x<\/em> + b) = x<\/em>2<\/sup> + (a + b) x<\/em> + ab<\/p>\n\n\n\n

Here, x<\/em> = 100, a = 3 and b = 7
103 \u00d7 107 = (100 + 3) (100 + 7) = (100)2<\/sup> + (3 + 7)10 + (3 \u00d7 7)
                 = 10000 + 100 + 21 
                 = 10121<\/p>\n\n\n\n

(ii) 95 \u00d7 96 = (90 + 5) (90 + 4)
Using identity, (x <\/em>+ a) (x<\/em> + b) = x<\/em>2<\/sup> + (a + b) x<\/em> + ab 
Here, x<\/em> = 90, a = 5 and b = 4
95 \u00d7 96 = (90 + 5) (90 + 4) = 902<\/sup> + 90(5 + 6) + (5 \u00d7 6) <\/p>\n\n\n\n

             = 8100 + (11 \u00d7 90) + 30
             = 8100 + 990 + 30 = 9120
(iii) 104 \u00d7 96 = (100 + 4) (100 – 4)
Using identity, (x <\/em>+ y<\/em>) (x<\/em> –y<\/em>) = x<\/em>2<\/sup> – y<\/em>2<\/sup>
Here, x<\/em> = 100 and y<\/em> = 4
104 \u00d7 96 = (100 + 4) (100 – 4) = (100)2<\/sup> -(4)2 <\/sup>= 10000 – 16 = 9984 <\/p>\n\n\n\n

3. Factorise the following using appropriate identities:<\/strong>
   (i) 9x<\/em>2<\/sup> + 6xy<\/em> + y<\/em>2                 <\/sup>(ii) 4y<\/em>2<\/sup> – 4y<\/em> + 1              (iii) x2 <\/sup><\/em>– y<\/em>2<\/sup>\/100<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 9x2<\/sup> + 6xy + y2  <\/sup>= (3x) 2<\/sup> + (2\u00d73x\u00d7y) + y2<\/sup><\/em>
Using identity, (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
Here, a = 3x<\/em> and b = y<\/em>
9x2<\/sup> + 6xy + y2  <\/sup>= (3x) 2<\/sup> + (2\u00d73x\u00d7y) + y2 <\/sup>= (3x + y)2 <\/sup>=(3x + y) (3x + y)<\/em><\/p>\n\n\n\n

(ii) 4y2<\/sup> – 4y + 1 = (2y)2<\/sup> – (2\u00d72y\u00d71) + 12<\/sup><\/em>
Using identity, (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>
Here, a = 2y<\/em> and b = 1
4y2<\/sup> – 4y + 1 = (2y)2<\/sup> – (2\u00d72y\u00d71) + 12 <\/sup>= (2y – 1)2 <\/sup>=(2y – 1) (2y – 1)<\/em><\/p>\n\n\n\n

(iii) x2 <\/sup>– y2<\/sup>\/100 = x2 <\/sup>– (y\/10)2<\/sup><\/em>
Using identity, a2<\/sup> – b2<\/sup> = (a + b) (a – b)
Here, a = x<\/em> and b = (y<\/em>\/10)
x2 <\/sup>– y2<\/sup>\/100 = x2 <\/sup>– (y\/10)2 <\/sup>= (x- y\/10) (x+ y\/10)<\/em><\/p>\n\n\n\n

Page No: 49
4. Expand each of the following, using suitable identities:<\/strong>
    (i) (x<\/em> + 2y<\/em> + 4z<\/em>)2<\/sup>                     (ii) (2x<\/em> \u2013 y<\/em> + z<\/em>)2<\/sup>                    (iii) (\u20132x<\/em> + 3y<\/em> + 2z<\/em>)2<\/sup>
   (iv) (3a<\/em> \u2013 7b<\/em> \u2013 c<\/em>)2                       <\/sup>  (v) (\u20132x<\/em> + 5y<\/em> \u2013 3z<\/em>)2                   <\/sup>(vi) [1\/4 a<\/em> – 1\/2 b<\/em> + 1]2<\/sup> <\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) (x<\/em> + 2y<\/em> + 4z<\/em>)2<\/sup>
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca 
Here, a = x<\/em>, b = 2y <\/em>and c = 4z<\/em>
(x<\/em> + 2y<\/em> + 4z<\/em>)2 <\/sup>= x2<\/sup> + (2y)2<\/sup> + (4z)2<\/sup> + (2\u00d7x\u00d72y) + (2\u00d72y\u00d74z) + (2\u00d74z\u00d7x)<\/em>
                      = <\/em>x2<\/sup> + 4y2<\/sup> + 16z2<\/sup> + 4xy + 16yz + 8xz<\/em><\/p>\n\n\n\n

(ii)  (2x<\/em> \u2013 y<\/em> + z<\/em>)2<\/sup>
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca 
Here, a = 2x<\/em>, b = –y <\/em>and c = z<\/em>
(2x \u2013 y + z)2 <\/sup><\/em>= (2x)2<\/sup> + (-y)2<\/sup> + z2<\/sup> + (2<\/em>\u00d72x<\/em>\u00d7-y<\/em>) + (2<\/em>\u00d7-y<\/em>\u00d7z)<\/em> + (2<\/em>\u00d7z<\/em>\u00d72x)<\/em>
                     = 4x2<\/sup> + y2<\/sup> + z2<\/sup> – 4<\/em>x<\/em>y<\/em> – 2<\/em>y<\/em>z<\/em> + 4<\/em>xz<\/em><\/p>\n\n\n\n

(iii) (\u20132x<\/em> + 3y<\/em> + 2z<\/em>)2<\/sup> 
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca 
Here, a = -2x<\/em>, b = 3y <\/em>and c = 2z<\/em>
(\u20132x + 3y + 2z)2 <\/sup><\/em>= (-2x)2<\/sup> + (3y)2<\/sup> + (2z)2<\/sup> + (2<\/em>\u00d7-2x<\/em>\u00d73y<\/em>) + (2<\/em>\u00d73y<\/em>\u00d72z)<\/em> + (2<\/em>\u00d72z<\/em>\u00d7-2x)<\/em>
                     = 4x2<\/sup> + 9y2<\/sup> + 4z2<\/sup> – 12<\/em>x<\/em>y<\/em> + 12<\/em>y<\/em>z<\/em> – 8<\/em>xz<\/em><\/p>\n\n\n\n

(iv) (3a<\/em> \u2013 7b<\/em> \u2013 c<\/em>)2<\/sup>
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca 
Here, a = 3a<\/em>, b = -7band c = -c
(3a \u2013 7b \u2013 c)2<\/sup><\/em>= (3a)2<\/sup> + (-7b)2<\/sup> + (-c)2<\/sup> + (2<\/em>\u00d73a<\/em>\u00d7-7b<\/em>) + (2<\/em>\u00d7-7b<\/em>\u00d7-c)<\/em> + (2<\/em>\u00d7-c<\/em>\u00d73a)<\/em>
                     = 9a2<\/sup> + 49b2<\/sup> + c2<\/sup> – 42<\/em>ab<\/em> + 14bc<\/em> – 6ac<\/em><\/p>\n\n\n\n

(v) (\u20132x<\/em> + 5y<\/em> \u2013 3z<\/em>)2<\/sup>  
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca 
Here, a = -2x<\/em>, b = 5y <\/em>and c = -3z<\/em>
(\u20132x + 5y \u2013 3z)2<\/sup><\/em>= (-2x)2<\/sup> + (5y)2<\/sup> + (-3z)2<\/sup> + (2<\/em>\u00d7-2x<\/em>\u00d75y<\/em>) + (2<\/em>\u00d75y<\/em>\u00d7-3z)<\/em> + (2<\/em>\u00d7-3z<\/em>\u00d7-2x)<\/em>
                     = 4x2<\/sup> + 25y2<\/sup> + 9z2<\/sup> – 20<\/em>x<\/em>y<\/em> – 30<\/em>y<\/em>z<\/em> + 12<\/em>xz<\/em><\/p>\n\n\n\n

(vi) [1\/4 a<\/em> – 1\/2 b<\/em> + 1]2<\/sup> 
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca 
Here, a = 1\/4 a<\/em>, b = -1\/2 band c = 1
[1\/4 a – 1\/2 b + 1]2<\/sup><\/em>= (1\/4 a)2<\/sup> + (-1\/2 b)2<\/sup> + 12<\/sup> + (2<\/em>\u00d71\/4 a<\/em>\u00d7-1\/2 b<\/em>) + (2<\/em>\u00d7-1\/2 b<\/em>\u00d71)<\/em> + (2<\/em>\u00d71<\/em>\u00d71\/4 a)<\/em>
                                = 1\/16 a2<\/sup> + 1\/4 b2<\/sup> + 1 – 1\/4 <\/em>ab<\/em> – b<\/em> + 1\/2 a<\/em><\/p>\n\n\n\n

5. Factorise:<\/strong>
(i) 4x2<\/sup><\/em> + 9y2<\/sup><\/em> + 16z2<\/sup><\/em> + 12xy<\/em> – 24yz<\/em> – 16xz<\/em>
(ii) 2x2<\/sup><\/em> + y2<\/sup><\/em> + 8z2<\/sup><\/em> – 2\u221a2 xy<\/em> + 4\u221a2 yz<\/em> – 8xz<\/em>
Answer<\/strong>
(i) 4x2<\/sup><\/em> + 9y2<\/sup><\/em> + 16z2<\/sup><\/em> + 12xy<\/em> – 24yz<\/em> – 16xz<\/em>
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup> + 12xy – 24yz – 16xz<\/em>
= (2x)2<\/sup> + (3y)2<\/sup> + (-4z)2<\/sup> + (2\u00d72x\u00d73y) + (2\u00d73y\u00d7-4z) + (2\u00d7-4z\u00d72x)<\/em>
= (2x + 3y – 4z)2<\/sup><\/em>
= (2x + 3y – 4z) (2x + 3y – 4z)<\/em><\/p>\n\n\n\n

(ii) 2x2<\/sup><\/em> + y2<\/sup><\/em> + 8z2<\/sup><\/em> – 2\u221a2 xy<\/em> + 4\u221a2 yz<\/em> – 8xz<\/em>
Using identity, (a + b + c<\/em>)2 <\/sup>= a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
2x2<\/sup> + y2<\/sup> + 8z2<\/sup> – 2\u221a2 xy<\/em> + 4\u221a2 yz – 8xz<\/em> 
= (-\u221a2x)2<\/sup> + (y)2<\/sup> + (2\u221a2z)2<\/sup> + (2\u00d7-\u221a2x\u00d7y) + (2\u00d7y\u00d72\u221a2z) + (2\u00d72\u221a2z\u00d7-\u221a2x)<\/em>
= <\/em>(<\/em>-\u221a2x + y + <\/em>2\u221a2z)2<\/sup><\/em>
= <\/em>(-\u221a2x + y + 2\u221a2z) <\/em>(<\/em>-\u221a2x + y + <\/em>2\u221a2z)<\/em><\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

6. Write the following cubes in expanded form:<\/strong>
    (i) (2x + 1)3<\/sup>                 (ii) (2a \u2013 3b)3     <\/sup>           (iii) [3\/2 x<\/em> + 1]3<\/sup>           (iv) [x<\/em> – 2\/3 y<\/em>]3<\/sup><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i)(2x<\/em> + 1)3<\/sup>
Using identity, (a + b)3 <\/sup>= a3<\/sup> + b3<\/sup> + 3ab(a + b)
(2x + 1)3 <\/sup>= (2x)3<\/sup> + 13<\/sup> +<\/em> (3<\/em>\u00d7<\/em>2x<\/em>\u00d71)(2x + 1)<\/em>
= 8<\/em>x3<\/sup> + 1 + 6x(2x + 1)<\/em>
= <\/em>8<\/em>x3<\/sup> + 12<\/em>x2<\/sup> + <\/em>6x<\/em> + 1<\/em><\/p>\n\n\n\n

(ii)(2a \u2013 3b)3<\/sup>
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3ab(a – b) 
(2a \u2013 3b)3 <\/sup><\/em>= (2a)3<\/sup> – (3b)3<\/sup> – (3<\/em>\u00d72a<\/em>\u00d73b)(2a – 3b)<\/em>
= 8<\/em>a3<\/sup> – 27b3<\/sup> – 18a<\/em>b(2a – 3b)<\/em>
= <\/em>8<\/em>a3<\/sup> – 27b3<\/sup> – 36<\/em>a2<\/sup><\/em>b + 54ab2<\/sup><\/em><\/p>\n\n\n\n

(iii) [3\/2 x<\/em> + 1]3<\/sup> 
Using identity, (a + b)3 <\/sup>= a3<\/sup> + b3<\/sup> + 3ab(a + b)
[3\/2 x + 1]3 <\/sup><\/em>= (3\/2 x)3<\/sup> + 13<\/sup> +<\/em> (3<\/em>\u00d7<\/em>3\/2 x<\/em>\u00d71)(3\/2 x + 1)<\/em>
= 27\/8 <\/em>x3 <\/sup>+ 1 +<\/em> 9\/2 x(3\/2 x + 1)<\/em>
= <\/em>27\/8 <\/em>x3 <\/sup>+ 1 +<\/em> 27\/4 <\/em>x2<\/sup><\/em> + 9\/2 x<\/em>
= <\/em>27\/8 <\/em>x3 <\/sup>+<\/em> 27\/4 <\/em>x2<\/sup><\/em> + 9\/2 x <\/em>+ 1<\/em><\/p>\n\n\n\n

(iv) [x<\/em> – 2\/3 y<\/em>]3<\/sup>
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3ab(a – b)
[x – 2\/3 y]3<\/sup><\/em> = (x)3<\/sup> – (2\/3 y)3<\/sup> – (3\u00d7x\u00d72\/3 y)(x – 2\/3 y)<\/em>
= x3<\/sup> – 8\/27y3<\/sup> – 2xy(x – 2\/3 y)<\/em>
= x3<\/sup> – 8\/27y3<\/sup> – 2x2<\/sup><\/em>y + 4\/3xy2<\/sup><\/em>

7. Evaluate the following using suitable identities: <\/strong>
    (i) (99)3<\/sup><\/em>            (ii) (102)3<\/sup><\/em>             (iii) (998)3<\/sup><\/em><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) (99)3<\/sup><\/em> = (100 – 1)3<\/sup><\/em>
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3ab(a – b) 
(100 – 1)3 <\/sup><\/em>= (100)3<\/sup> – 13<\/sup> – (3<\/em>\u00d7100<\/em>\u00d71)(100 – 1)<\/em>
= 1000000 – 1 – 300(100 – 1)<\/em>
=<\/em> 1000000 – 1 – 30000 + 300<\/em>
= 970299<\/em><\/p>\n\n\n\n

(ii) (102)3<\/sup> = (100 + 2)3<\/sup>
Using identity, (a + b)3 <\/sup>= a3<\/sup> + b3<\/sup> + 3ab(a + b)
(100 + 2)3 <\/sup><\/em>= (100)3<\/sup> + 23<\/sup> + (3<\/em>\u00d7100<\/em>\u00d72)(100 + 2)<\/em>
= 1000000 + 8 + 600(100 + 2)<\/em>
=<\/em> 1000000 <\/em>+ 8 + 60000 + 1200<\/em>
= 1061208<\/em><\/p>\n\n\n\n

(iii) (998)3<\/sup>
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3ab(a – b) 
(1000 – 2)3 <\/sup><\/em>= (1000)3<\/sup> – 23<\/sup> – (3<\/em>\u00d71000<\/em>\u00d72)(1000 – 2)<\/em>
= 100000000 – 8 – 6000(1000 – 2)<\/em>
=<\/em> 100000000 – 8- 600000 + 12000<\/em>
= 994011992<\/em><\/p>\n\n\n\n

8. Factorise each of the following:<\/strong>
(i) 8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup>                           (ii) 8a3<\/sup> – b3<\/sup> – 12a2<\/sup>b + 6ab2<\/sup>
(iii) 27 – 125a3<\/sup> – 135a + 225a2<\/sup>                      (iv) 64a3<\/sup> – 27b3<\/sup> – 144a2<\/sup>b + 108ab2<\/sup>
(v) 27p3<\/sup> – 1\/216 – 9\/2 p2<\/sup> + 1\/4 p
Answer<\/strong><\/p>\n\n\n\n

(i) 8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup><\/em>
Using identity, (a + b)3 <\/sup>= a3<\/sup> + b3<\/sup> + 3a2<\/sup>b + 3ab2<\/sup>
8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup> <\/em>
=<\/em> (2a)3<\/sup> + b3<\/sup> +<\/em> 3(<\/em>2a)<\/em>2<\/sup><\/em>b +<\/em> 3(2a<\/em>)(b<\/em>)2<\/sup><\/em>
= (2a + b)3<\/sup><\/em>
= (2a + b)(2a + b)(2a + b)<\/em><\/p>\n\n\n\n

(ii) 8a3<\/sup> – b3<\/sup> – 12a2<\/sup>b + 6ab2<\/sup>
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3a2<\/sup>b + 3ab2<\/sup>
8a3<\/sup> – b3<\/sup> – 12a2<\/sup>b + 6ab2<\/sup><\/em>=<\/em> (2a)3<\/sup> – b3<\/sup> –<\/em> 3(<\/em>2a)<\/em>2<\/sup><\/em>b +<\/em> 3(2a<\/em>)(b<\/em>)2<\/sup><\/em>
= (2a – b)3<\/sup><\/em>
= (2a – b)(2a – b)(2a – b)<\/em><\/p>\n\n\n\n

(iii) 27 – 125a3<\/sup> – 135a + 225a2<\/sup>
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3a2<\/sup>b + 3ab2<\/sup>
27 – 125a3<\/sup> – 135a + 225a2<\/sup><\/em>=<\/em> 33<\/sup> – (5a)3<\/sup> –<\/em> 3(<\/em>3)<\/em>2<\/sup>(5a) +<\/em> 3(3<\/em>)(<\/em>5a<\/em>)2<\/sup><\/em>
= (3 – 5a)3<\/sup><\/em>
= <\/em>(3 – 5a)<\/em>(3 – 5a)<\/em>(3 – 5a)<\/em><\/p>\n\n\n\n

(iv) 64a3<\/sup> – 27b3<\/sup> – 144a2<\/sup>b + 108ab2<\/sup>
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3a2<\/sup>b + 3ab2<\/sup>
64a3<\/sup> – 27b3<\/sup> – 144a2<\/sup>b + 108ab2<\/sup><\/em>=<\/em> (4a)3<\/sup> – (3b)3<\/sup> –<\/em> 3(<\/em>4a)<\/em>2<\/sup>(3b) + 3<\/em>(4a<\/em>)(<\/em>3b<\/em>)2<\/sup><\/em>
= (4a – 3b)3<\/sup><\/em>
= (4a – 3b)(4a – 3b)(4a – 3b)<\/em><\/p>\n\n\n\n

(v) 27p3<\/sup> – 1\/216 – 9\/2 p2<\/sup> + 1\/4 p 
Using identity, (a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3a2<\/sup>b + 3ab2<\/sup>
27p3<\/sup> – 1\/216 – 9\/2 p2<\/sup> + 1\/4 p<\/em>
= <\/em>(3p)3<\/sup> – (1\/6)3<\/sup> –<\/em> 3(<\/em>3p)<\/em>2<\/sup>(1\/6) + 3<\/em>(3p<\/em>)(<\/em>1\/6<\/em>)2<\/sup><\/em>
= (3p – 1\/6)3<\/sup><\/em>
= (3p – 1\/6)(3p – 1\/6)(3p – 1\/6)<\/em><\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

9. Verify : <\/strong>(i) x3<\/sup><\/em> + y3<\/sup><\/em> = (x<\/em> + y<\/em>) (x2<\/sup><\/em> – xy<\/em> + y2<\/sup><\/em>)             (ii) x3<\/sup><\/em> – y3<\/sup><\/em> = (x<\/em> – y<\/em>) (x2<\/sup><\/em> + xy<\/em> + y2<\/sup><\/em>)<\/p>\n\n\n\n

Answer<\/strong>
(i) x<\/em>3<\/sup><\/em> + y<\/em>3<\/sup><\/em> = (x<\/em> + y<\/em>) (x<\/em>2<\/sup><\/em> – xy<\/em> + y2<\/sup><\/em>)
We know that, 
(x + y)3 <\/sup>= x3<\/sup> + y3<\/sup> +<\/em> 3xy(x + y)<\/em>
\u21d2 x3<\/sup> + y3 <\/sup>= (x + y)3 <\/sup>-3xy(x + y)<\/em>
\u21d2 x3<\/sup> + y3 <\/sup>= (x + y)[(x + y)2<\/sup> -3xy]                  <\/em>{Taking (x+y) common}<\/em>
\u21d2 x3<\/sup> + y3 <\/sup>= (x + y)[(x<\/em>2<\/sup> + y2 <\/sup>+ 2xy) -3xy] <\/em>
\u21d2 x3<\/sup> + y3 <\/sup>= (x + y)(x2<\/sup> + y2 <\/sup>– xy)<\/em><\/p>\n\n\n\n

(ii) x<\/em>3<\/sup><\/em> – y<\/em>3<\/sup><\/em> = (x<\/em> – y<\/em>) (x<\/em>2<\/sup><\/em> + xy<\/em> + y2<\/sup><\/em> )
We know that, 
(x – y)3 <\/sup>= x3<\/sup> – y3<\/sup> –<\/em> 3xy(x – y)<\/em>
\u21d2 x3<\/sup> – y3 <\/sup>= (x – y)3 <\/sup>+3xy(x – y)<\/em>
\u21d2 x3<\/sup> + y3 <\/sup>= (x – y)[(x – y)2<\/sup> +3xy]                     {Taking (x-y) common}<\/em>
\u21d2 x3<\/sup> + y3 <\/sup>= (x – y)[(x<\/em>2<\/sup> + y2 <\/sup>– 2xy) +3xy] <\/em>
\u21d2 x3<\/sup> + y3 <\/sup>= (x + y)(x2<\/sup> + y2 <\/sup>+ xy)<\/em><\/p>\n\n\n\n

10. Factorise each of the following:<\/strong>
      (i) 27y3<\/sup> + 125z3<\/sup>                     (ii) 64m3<\/sup> – 343n3<\/sup><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 27y3<\/sup> + 125z3<\/sup>
Using identity, x<\/em>3<\/sup><\/em> + y<\/em>3<\/sup><\/em> = (x<\/em> + y<\/em>) (x<\/em>2<\/sup><\/em> – xy<\/em> + y2<\/sup><\/em>)
27y3<\/sup> + 125z3<\/sup> =<\/em> (3y)<\/em>3<\/sup> + (5z)3<\/sup><\/em>
= <\/em>(3y<\/em> + 5z<\/em>) {(3y)2<\/sup><\/em> – (3y)(5z)<\/em> + (5z)2<\/sup><\/em>}
=<\/em> (3y + 5z) (9y2<\/sup> – 15yz + 25z)2<\/sup> <\/em><\/p>\n\n\n\n

(ii) 64m3<\/sup> – 343n3<\/sup>
Using identity, x3<\/sup> – y3<\/sup> = (x – y) (x2<\/sup> + xy + y2<\/sup> ) <\/em>
64m3<\/sup> – 343n3<\/sup> =<\/em> (4m)<\/em>3<\/sup> – (7n)3<\/sup><\/em>
= <\/em>(4m<\/em> + 7n<\/em>) {(4m)2<\/sup><\/em> + (4m)(7n)<\/em> + (7n)2<\/sup><\/em>}
=<\/em> (<\/em>4m + 7n) (16m2<\/sup> + 28mn + 49n)2<\/sup> <\/em><\/p>\n\n\n\n

11. Factorise : <\/strong>27x3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 9xyz<\/em>
Answer<\/strong><\/p>\n\n\n\n

27x3<\/sup> + y3<\/sup> + z3<\/sup> – 9xyz = (3x)3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 3\u00d7<\/em>3xyz<\/em>
Using identity,x3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 3xyz = (x + y + z)(<\/em>x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – xz)<\/em>
27x3<\/sup> + y3<\/sup> + z3<\/sup> – 9xyz<\/em>
= <\/em>(3x + y + z) {(3<\/em>x)2<\/sup> + y2<\/sup> + z2<\/sup> – 3xy – yz – 3xz}<\/em>
= <\/em>(3x + y + z) (9<\/em>x2<\/sup> + y2<\/sup> + z2<\/sup> – 3xy – yz – 3xz)<\/em><\/p>\n\n\n\n

12. Verify that:<\/strong> x3<\/sup><\/em> + y3<\/sup> + z3<\/sup> – <\/em>3xyz = <\/em>1\/2(x + y + z<\/em>) [(x – y<\/em>)2 <\/sup><\/em>+ (y – z<\/em>)2 <\/sup><\/em>+(z – x<\/em>)2<\/sup><\/em>]<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

We know that,
x3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 3xyz = (x + y + z)(<\/em>x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – xz)<\/em>
\u21d2 x3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 3xyz = 1\/2<\/em>\u00d7<\/em>(x + y + z) 2(x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – xz)<\/em>
= 1\/2<\/em>(x + y + z) (2<\/em>x2<\/sup> + 2y2<\/sup> + 2z2<\/sup> – 2xy – 2yz – 2xz)<\/em> 
= 1\/2<\/em>(x + y + z) [(<\/em>x2<\/sup> + y2<\/sup> -2xy) + (<\/em>y2 <\/sup><\/em>+ z2<\/sup> – 2yz) + (<\/em>x2<\/sup> + <\/em>z2 <\/sup><\/em>– 2xz)] 
= 1\/2(x + y + z) [(x – y)2 <\/sup>+ (y – z)2 <\/sup>+(z – x)2<\/sup>]<\/em><\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

13. If x + y + z = <\/em>0, show that x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz.<\/em><\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

We know that,
x3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 3xyz = (x + y + z)(<\/em>x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – xz)<\/em>
Now put <\/em>(x + y + z) = 0,  <\/em>
x3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 3xyz = (0)(<\/em>x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – xz) <\/em>
\u21d2 <\/em>x3<\/sup><\/em> + y3<\/sup><\/em> + z3<\/sup><\/em> – 3xyz = 0<\/em><\/p>\n\n\n\n

14. Without actually calculating the cubes, find the value of each of the following:<\/strong>
     (i) (-12)3<\/sup><\/em> + (7)3<\/sup><\/em> + (5)3<\/sup><\/em>
     (ii) (28)3<\/sup><\/em> + (\u201315)3<\/sup><\/em> + (-13)3<\/sup><\/em><\/p>\n\n\n\n

Answer<\/strong>
(i) (-12)3<\/sup><\/em> + (7)3<\/sup><\/em> + (5)3<\/sup><\/em>
Let x<\/em> =-12, y<\/em> = 7 and z<\/em> = 5
We observed that, x + y + z<\/em> = -12 + 7 + 5 = 0
We know that if,
x + y + z = <\/em>0, then x3<\/sup> +<\/em> y3<\/sup> + z3<\/sup> = 3xyz<\/em>
(-12)3<\/sup> + (7)3<\/sup> + (5)3<\/sup> = 3(-12)(7)(5) = -1260<\/em><\/p>\n\n\n\n

(ii)(28)3<\/sup><\/em> + (\u201315)3<\/sup><\/em> + (-13)3<\/sup><\/em>
Let x<\/em> =28, y<\/em> = -15 and z<\/em> = -13
We observed that, x + y + z<\/em> = 28 – 15 – 13 = 0
We know that if,
x + y + z = <\/em>0, then x3<\/sup> +<\/em> y3<\/sup> + z3<\/sup> = 3xyz<\/em>
(28)3<\/sup> + (\u201315)3<\/sup> + (-13)3<\/sup> = 3(28)(-15)(-13) = 16380<\/em><\/p>\n\n\n\n

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:<\/strong>
(i) Area : 25a2<\/sup><\/em> – 35a + 12
(ii) Area : 35 y2<\/sup><\/em> + 13y<\/em> – 12
Answer<\/strong><\/p>\n\n\n\n

(i) Area : 25a2<\/sup> – 35a + 12
Since, area is product of length and breadth therefore by factorizing the given area, we can know the length and breadth of rectangle.
25a2<\/sup> – 35a + 12<\/em>
= <\/em>25a2<\/sup> – 15a -20a + 12<\/em>
= 5a(5a – 3) – 4(5a – 3)<\/em>
= <\/em>(5a – 4)<\/em>(5a – 3)<\/em>
Possible expression for length = 5a – 4<\/em>
Possible expression for breadth = 5a – 3<\/em><\/p>\n\n\n\n

(ii)Area : 35 y2<\/sup><\/em> + 13y<\/em> – 12
35 y2<\/sup> + 13y – 12<\/em>
= 35y2<\/sup> – 15y + 28y – 12<\/em>
= 5y(7y – 3) + 4(7y – 3)<\/em>
= (5y + 4)(7y – 3)<\/em>
Possible expression for length = (5y + 4)<\/em>
Possible expression for breadth = (7y – 3)<\/em><\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

Page No: 50<\/p>\n\n\n\n

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?(i) Volume :<\/strong> 3x2<\/sup> – <\/em>12x<\/em>
(ii) Volume : 12ky2<\/sup> + <\/em>8ky – <\/em>20k<\/em><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Volume : 3x2<\/sup><\/em> – 12x <\/em>
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid.
3x2<\/sup> – 12x<\/em>
= 3x(x – 4)<\/em>
Possible expression for length = <\/em>3<\/em>
Possible expression for breadth = <\/em>x<\/em>
Possible expression for height = <\/em>(x – 4)<\/em><\/p>\n\n\n\n

(ii) Volume : 12k<\/em>y2<\/sup> + <\/em>8ky – <\/em>20k<\/em>
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid.
12ky2<\/sup> + 8ky – 20k<\/em>
= 4k(<\/em>3y2<\/sup> + 2y – 5)<\/em>
= 4k(<\/em>3y2<\/sup> +5y – 3y – 5)<\/em>
= <\/em>4k[y(<\/em>3y +5) – 1(3y + 5)]<\/em>
= <\/em>4k (<\/em>3y +5) (y – 1)<\/em>
Possible expression for length = 4k<\/em>
Possible expression for breadth = <\/em>(<\/em>3y +5)<\/em>
Possible expression for height = <\/em>(y – 1)<\/em><\/p>\n\n\n\n

We have already understood about algebraic expressions in previous class and in this chapter we will studying about a particular algebraic expression, Polynomial. What is Polynomial? An algebraic expression containing one or more than one terms is called Polynomial. We will study the Remainder Theorem, Factor Theorem and algebraic identities and their use in factorisation of polynomials.<\/p>\n\n\n\n

\u2022 Polynomials in one Variable: We will discuss linear, quadratic and cubic polynomial in this section briefly.<\/p>\n\n\n\n

\u2022 Zeroes of a Polynomial: You will know how to find the zero of the polynomial in given cases.<\/p>\n\n\n\n

\u2022 Remainder Theorem: Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x \u2013 a, then the remainder is p(a).<\/p>\n\n\n\n

\u2022 Factorisation of Polynomials: If p(x) is a polynomial of degree n > 1 and a is any real number, then (i) x \u2013 a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x \u2013 a is a factor of p(x). <\/p>\n\n\n\n

\u2022 Algebraic Identities: We will be solving questions of the exercise based on the given four identities:<\/p>\n\n\n\n

Identity I: (x + y)2<\/sup><\/em> = x2<\/sup><\/em> + 2xy + y2<\/sup><\/em><\/p>\n\n\n\n

Identity II: (x \u2013 y)2<\/sup><\/em> = x2<\/sup><\/em> \u2013 2xy + y2<\/sup><\/em><\/p>\n\n\n\n

Identity III : x2<\/sup><\/em> \u2013 y2<\/sup><\/em> = (x + y) (x \u2013 y)<\/p>\n\n\n\n

Identity IV : (x + a) (x + b) = x2<\/sup><\/em> + (a + b)x + ab<\/p>\n\n\n\n

You will find total 5 exercises in the whole chapter that are very helpful in building your basic concepts. Indcareer Schools experts have prepared exercisewise answers of every question for Chapter 2 Polynomials NCERT Solutions Class 9<\/strong> which you can find below.<\/p>\n\n\n\n

These NCERT Solutions for Class 9 Maths prepared by Indcareer Schools will help you in revising whole topics in less time and improving skills.<\/p>\n\n\n\n

NCERT Solutions for Class 9 Maths Chapters:<\/h4>\n\n\n\n

FAQ on <\/strong>Chapter 2 Polynomials<\/strong>
<\/p>\n\n\n\n

How can I download Chapter 2 Polynomials Class 9 NCERT Solutions PDF <\/h4>\n\n\n\n

Chapter 2 Polynomials Class 9 Maths NCERT Solutions PDF download is very useful in understanding the basic concepts embedded in the chapter. It is very essential to solve every question before moving further to any other supplementary books.<\/p>\n\n\n\n

What do you mean by Algebraic Expressions?<\/h4>\n\n\n\n

A combination of constants and variables, connected by some or all the basic operations +, \u2013, \u00d7, \u00f7 is called an algebraic expression. <\/p>\n\n\n\n

What is the coefficient of a zero polynomial?<\/h4>\n\n\n\n

The coefficient of a zero polynomial is 1.<\/p>\n\n\n\n

What is Biquadratic Polynomial?<\/h4>\n\n\n\n

A polynomial of degree 4 is called a biquadratic polynomial.<\/p>\n\n\n\n

NCERT 9th Maths Chapter 2, class 9 Maths Chapter 2 solutions<\/p>\n\n\n\n

NCERT Solutions for 9th Class Maths :Chapter 2: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 9th Class Maths :Chapter 2 Polynomials<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 9th Class Maths :Chapter 2 Polynomials PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 9 Maths :<\/strong><\/h4>\n\n\n\n