{"id":545436,"date":"2021-10-04T10:07:33","date_gmt":"2021-10-04T10:07:33","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545436"},"modified":"2021-10-05T09:15:21","modified_gmt":"2021-10-05T09:15:21","slug":"rd-sharma-solutions-for-class-9-maths-chapter-6-factorization-of-polynomials","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-6-factorization-of-polynomials\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 6\u2013Factorization Of Polynomials"},"content":{"rendered":"\n

Class 9: Maths Chapter 6 solutions. Complete Class 9 Maths Chapter 6 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 6\u2013Factorization Of Polynomials<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 6, Class 9 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.1 Page No: 6.2<\/h4>\n\n\n\n

Question 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) 3x2<\/sup> \u2013 4x + 15<\/strong><\/p>\n\n\n\n

(ii) y2<\/sup> + 2\u221a3<\/strong><\/p>\n\n\n\n

(iii) 3\u221ax + \u221a2x<\/strong><\/p>\n\n\n\n

(iv) x \u2013 4\/x<\/strong><\/p>\n\n\n\n

(v) x12<\/sup> + y3<\/sup> + t50<\/sup><\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 3x2<\/sup> \u2013 4x + 15<\/p>\n\n\n\n

It is a polynomial of x.<\/p>\n\n\n\n

(ii)<\/strong> y2<\/sup> + 2\u221a3<\/p>\n\n\n\n

It is a polynomial of y.<\/p>\n\n\n\n

(iii)<\/strong> 3\u221ax + \u221a2x<\/p>\n\n\n\n

It is not a polynomial since the exponent of 3\u221ax is a rational term.<\/p>\n\n\n\n

(iv)<\/strong> x \u2013 4\/x<\/p>\n\n\n\n

It is not a polynomial since the exponent of \u2013 4\/x is not a positive term.<\/p>\n\n\n\n

(v)<\/strong> x12<\/sup> + y3<\/sup> + t50<\/sup><\/p>\n\n\n\n

It is a three variable polynomial, x, y and t.<\/p>\n\n\n\n

Question 2: Write the coefficient of x2<\/sup> in each of the following:<\/strong><\/p>\n\n\n\n

(i) 17 \u2013 2x + 7x2<\/sup><\/strong><\/p>\n\n\n\n

(ii) 9 \u2013 12x + x3<\/sup><\/strong><\/p>\n\n\n\n

(iii) \u220f\/6 x2<\/sup> \u2013 3x + 4<\/strong><\/p>\n\n\n\n

(iv) \u221a3x \u2013 7<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 17 \u2013 2x + 7x2<\/sup><\/p>\n\n\n\n

Coefficient of x2<\/sup> = 7<\/p>\n\n\n\n

(ii)<\/strong> 9 \u2013 12x + x3<\/sup><\/p>\n\n\n\n

Coefficient of x2 <\/sup>=0<\/p>\n\n\n\n

(iii)<\/strong> \u220f\/6 x2<\/sup> \u2013 3x + 4<\/p>\n\n\n\n

Coefficient of x2 <\/sup>= \u220f\/6<\/p>\n\n\n\n

(iv)<\/strong> \u221a3x \u2013 7<\/p>\n\n\n\n

Coefficient of x2 <\/sup>= 0<\/p>\n\n\n\n

Question 3: Write the degrees of each of the following polynomials:<\/strong><\/p>\n\n\n\n

(i) 7x3<\/sup> + 4x2<\/sup> \u2013 3x + 12<\/strong><\/p>\n\n\n\n

(ii) 12 \u2013 x + 2x3<\/sup><\/strong><\/p>\n\n\n\n

(iii) 5y \u2013 \u221a2<\/strong><\/p>\n\n\n\n

(iv) 7<\/strong><\/p>\n\n\n\n

(v) 0<\/strong><\/p>\n\n\n\n

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

As we know, degree is the highest power in the polynomial<\/p>\n\n\n\n

(i)<\/strong> Degree of the polynomial 7x3<\/sup> + 4x2<\/sup> \u2013 3x + 12 is 3<\/p>\n\n\n\n

(ii)<\/strong> Degree of the polynomial 12 \u2013 x + 2x3<\/sup> is 3<\/p>\n\n\n\n

(iii)<\/strong> Degree of the polynomial 5y \u2013 \u221a2 is 1<\/p>\n\n\n\n

(iv)<\/strong> Degree of the polynomial 7 is 0<\/p>\n\n\n\n

(v)<\/strong> Degree of the polynomial 0 is undefined.<\/p>\n\n\n\n

Question 4: Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:<\/strong><\/p>\n\n\n\n

(i) x + x2<\/sup> + 4<\/strong><\/p>\n\n\n\n

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

(iii) 2x + x2<\/sup><\/strong><\/p>\n\n\n\n

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

(v) t2<\/sup> + 1<\/strong><\/p>\n\n\n\n

(vi) 7t4<\/sup> + 4t3<\/sup> + 3t \u2013 2<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> x + x2<\/sup> + 4: It is a quadratic polynomial as its degree is 2.<\/p>\n\n\n\n

(ii)<\/strong> 3x \u2013 2 : It is a linear polynomial as its degree is 1.<\/p>\n\n\n\n

(iii)<\/strong> 2x + x2<\/sup>: It is a quadratic polynomial as its degree is 2.<\/p>\n\n\n\n

(iv)<\/strong> 3y: It is a linear polynomial as its degree is 1.<\/p>\n\n\n\n

(v)<\/strong> t2<\/sup>+ 1: It is a quadratic polynomial as its degree is 2.<\/p>\n\n\n\n

(vi)<\/strong> 7t4<\/sup> + 4t3<\/sup> + 3t \u2013 2: It is a biquadratic polynomial as its degree is 4.<\/p>\n\n\n\n

\n\n\n\n

Exercise 6.2 Page No: 6.8<\/h4>\n\n\n\n

Question 1: If f(x) = 2x3<\/sup> \u2013 13x2<\/sup> + 17x + 12, find<\/strong><\/p>\n\n\n\n

(i) f (2)<\/strong><\/p>\n\n\n\n

(ii) f (-3)<\/strong><\/p>\n\n\n\n

(iii) f(0)<\/strong><\/p>\n\n\n\n

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

f(x) = 2x3<\/sup> \u2013 13x2<\/sup> + 17x + 12<\/p>\n\n\n\n

(i)<\/strong> f(2) = 2(2)3<\/sup> \u2013 13(2) 2<\/sup> + 17(2) + 12<\/p>\n\n\n\n

= 2 x 8 \u2013 13 x 4 + 17 x 2 + 12<\/p>\n\n\n\n

= 16 \u2013 52 + 34 + 12<\/p>\n\n\n\n

= 62 \u2013 52<\/p>\n\n\n\n

= 10<\/p>\n\n\n\n

(ii)<\/strong> f(-3) = 2(-3)3<\/sup> \u2013 13(-3) 2<\/sup> + 17 x (-3) + 12<\/p>\n\n\n\n

= 2 x (-27) \u2013 13 x 9 + 17 x (-3) + 12<\/p>\n\n\n\n

= -54 \u2013 117 -51 + 12<\/p>\n\n\n\n

= -222 + 12<\/p>\n\n\n\n

= -210<\/p>\n\n\n\n

(iii)<\/strong> f(0) = 2 x (0)3<\/sup> \u2013 13(0) 2<\/sup> + 17 x 0 + 12<\/p>\n\n\n\n

= 0-0 + 0+ 12<\/p>\n\n\n\n

= 12<\/p>\n\n\n\n

Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:<\/strong><\/p>\n\n\n\n

(i) f(x) = 3x + 1, x = \u22121\/3<\/strong><\/p>\n\n\n\n

(ii) f(x) = x2<\/sup> \u2013 1, x = 1,\u22121<\/strong><\/p>\n\n\n\n

(iii) g(x) = 3x2<\/sup> \u2013 2 , x = 2\/\u221a3 , \u22122\/\u221a3<\/strong><\/p>\n\n\n\n

(iv) p(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6 , x = 1, 2, 3<\/strong><\/p>\n\n\n\n

(v) f(x) = 5x \u2013 \u03c0, x = 4\/5<\/strong><\/p>\n\n\n\n

(vi) f(x) = x2<\/sup> , x = 0<\/strong><\/p>\n\n\n\n

(vii) f(x) = lx + m, x = \u2212m\/l<\/strong><\/p>\n\n\n\n

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

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

(i)<\/strong> f(x) = 3x + 1, x = \u22121\/3<\/p>\n\n\n\n

f(x) = 3x + 1<\/p>\n\n\n\n

Substitute x = \u22121\/3 in f(x)<\/p>\n\n\n\n

f( \u22121\/3) = 3(\u22121\/3) + 1<\/p>\n\n\n\n

= -1 + 1<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Since, the result is 0, so x = \u22121\/3 is the root of 3x + 1<\/p>\n\n\n\n

(ii)<\/strong> f(x) = x2<\/sup> \u2013 1, x = 1,\u22121<\/p>\n\n\n\n

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

Given that x = (1 , -1)<\/p>\n\n\n\n

Substitute x = 1 in f(x)<\/p>\n\n\n\n

f(1) = 12<\/sup> \u2013 1<\/p>\n\n\n\n

= 1 \u2013 1<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Now, substitute x = (-1) in f(x)<\/p>\n\n\n\n

f(-1) = (\u22121)2<\/sup> \u2013 1<\/p>\n\n\n\n

= 1 \u2013 1<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x2 <\/sup>\u2013 1<\/p>\n\n\n\n

(iii)<\/strong> g(x) = 3x2<\/sup> \u2013 2 , x = 2\/\u221a3 , \u22122\/\u221a3<\/p>\n\n\n\n

g(x) = 3x2<\/sup> \u2013 2<\/p>\n\n\n\n

Substitute x = 2\/\u221a3 in g(x)<\/p>\n\n\n\n

g(2\/\u221a3) = 3(2\/\u221a3)2<\/sup> \u2013 2<\/p>\n\n\n\n

= 3(4\/3) \u2013 2<\/p>\n\n\n\n

= 4 \u2013 2<\/p>\n\n\n\n

= 2 \u2260 0<\/p>\n\n\n\n

Now, Substitute x = \u22122\/\u221a3 in g(x)<\/p>\n\n\n\n

g(2\/\u221a3) = 3(-2\/\u221a3)2<\/sup> \u2013 2<\/p>\n\n\n\n

= 3(4\/3) \u2013 2<\/p>\n\n\n\n

= 4 \u2013 2<\/p>\n\n\n\n

= 2 \u2260 0<\/p>\n\n\n\n

Since, the results when x = 2\/\u221a3 and x = \u22122\/\u221a3) are not 0. Therefore (2\/\u221a3 , \u22122\/\u221a3 ) are not zeros of 3x2<\/sup>\u20132.<\/p>\n\n\n\n

(iv)<\/strong> p(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6 , x = 1, 2, 3<\/p>\n\n\n\n

p(1) = 13<\/sup> \u2013 6(1)2<\/sup> + 11x 1 \u2013 6 = 1 \u2013 6 + 11 \u2013 6 = 0<\/p>\n\n\n\n

p(2) = 23<\/sup> \u2013 6(2)2<\/sup> + 11\u00d72 \u2013 6 = 8 \u2013 24 + 22 \u2013 6 = 0<\/p>\n\n\n\n

p(3) = 33<\/sup> \u2013 6(3)2<\/sup> + 11\u00d73 \u2013 6 = 27 \u2013 54 + 33 \u2013 6 = 0<\/p>\n\n\n\n

Therefore, x = 1, 2, 3 are zeros of p(x).<\/p>\n\n\n\n

(v)<\/strong> f(x) = 5x \u2013 \u03c0, x = 4\/5<\/p>\n\n\n\n

f(4\/5) = 5 x 4\/5 \u2013 \u03c0 = 4 \u2013 \u03c0 \u2260 0<\/p>\n\n\n\n

Therefore, x = 4\/5 is not a zeros of f(x).<\/p>\n\n\n\n

(vi)<\/strong> f(x) = x2<\/sup> , x = 0<\/p>\n\n\n\n

f(0) = 02<\/sup> = 0<\/p>\n\n\n\n

Therefore, x = 0 is a zero of f(x).<\/p>\n\n\n\n

(vii)<\/strong> f(x) = lx + m, x = \u2212m\/l<\/p>\n\n\n\n

f(\u2212m\/l) = l x \u2212m\/l + m = -m + m = 0<\/p>\n\n\n\n

Therefore, x = \u2212m\/l is a zero of f(x).<\/p>\n\n\n\n

(viii)<\/strong> f(x) = 2x + 1, x = \u00bd<\/p>\n\n\n\n

f(1\/2) = 2x 1\/2 + 1 = 1 + 1 = 2 \u2260 0<\/p>\n\n\n\n

Therefore, x = \u00bd is not a zero of f(x).<\/p>\n\n\n\n

\n\n\n\n

Exercise 6.3 Page No: 6.14<\/h4>\n\n\n\n

In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division : (1 \u2013 8)<\/strong><\/p>\n\n\n\n

Question 1: f(x) = x3<\/sup> + 4x2<\/sup> \u2013 3x + 10, g(x) = x + 4<\/strong><\/p>\n\n\n\n

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

f(x) = x3<\/sup> + 4x2<\/sup> \u2013 3x + 10, g(x) = x + 4<\/p>\n\n\n\n

Put g(x) =0<\/p>\n\n\n\n

\u21d2 x + 4 = 0 or x = -4<\/p>\n\n\n\n

Remainder = f(-4)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(-4) = (-4)3<\/sup> + 4(-4)2<\/sup> \u2013 3(-4) + 10 = -64 + 64 + 12 + 10 = 22<\/p>\n\n\n\n

Actual Division:<\/strong><\/p>\n\n\n\n

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

Question 2: f(x) = 4x4<\/sup> \u2013 3x3<\/sup> \u2013 2x2<\/sup> + x \u2013 7, g(x) = x \u2013 1<\/strong><\/p>\n\n\n\n

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

f(x) = 4x4<\/sup> \u2013 3x3<\/sup> \u2013 2x2<\/sup> + x \u2013 7<\/p>\n\n\n\n

Put g(x) =0<\/p>\n\n\n\n

\u21d2 x \u2013 1 = 0 or x = 1<\/p>\n\n\n\n

Remainder = f(1)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(1) = 4(1)4<\/sup> \u2013 3(1)3<\/sup> \u2013 2(1)2<\/sup> + (1) \u2013 7 = 4 \u2013 3 \u2013 2 + 1 \u2013 7 = -7<\/p>\n\n\n\n

Actual Division:<\/strong><\/p>\n\n\n\n

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

Question 3: f(x) = 2x4<\/sup> \u2013 6X3<\/sup> + 2x2<\/sup> \u2013 x + 2, g(x) = x + 2<\/strong><\/p>\n\n\n\n

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

f(x) = 2x4<\/sup> \u2013 6X3<\/sup> + 2x2<\/sup> \u2013 x + 2, g(x) = x + 2<\/p>\n\n\n\n

Put g(x) = 0<\/p>\n\n\n\n

\u21d2 x + 2 = 0 or x = -2<\/p>\n\n\n\n

Remainder = f(-2)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(-2) = 2(-2)4<\/sup> \u2013 6(-2)3<\/sup> + 2(-2)2<\/sup> \u2013 (-2) + 2 = 32 + 48 + 8 + 2 + 2 = 92<\/p>\n\n\n\n

Actual Division:<\/strong><\/p>\n\n\n\n

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

Question 4: f(x) = 4x3<\/sup> \u2013 12x2<\/sup> + 14x \u2013 3, g(x) = 2x \u2013 1<\/strong><\/p>\n\n\n\n

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

f(x) = 4x3<\/sup> \u2013 12x2<\/sup> + 14x \u2013 3, g(x) = 2x \u2013 1<\/p>\n\n\n\n

Put g(x) =0<\/p>\n\n\n\n

\u21d2 2x -1 =0 or x = 1\/2<\/p>\n\n\n\n

Remainder = f(1\/2)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(1\/2) = 4(1\/2)3<\/sup> \u2013 12(1\/2)2<\/sup> + 14(1\/2) \u2013 3 = \u00bd \u2013 3 + 7 \u2013 3 = 3\/2<\/p>\n\n\n\n

Actual Division:<\/strong><\/p>\n\n\n\n

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

Question 5: f(x) = x3<\/sup> \u2013 6x2<\/sup> + 2x \u2013 4, g(x) = 1 \u2013 2x<\/strong><\/p>\n\n\n\n

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

f(x) = x3<\/sup> \u2013 6x2<\/sup> + 2x \u2013 4, g(x) = 1 \u2013 2x<\/p>\n\n\n\n

Put g(x) = 0<\/p>\n\n\n\n

\u21d2 1 \u2013 2x = 0 or x = 1\/2<\/p>\n\n\n\n

Remainder = f(1\/2)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(1\/2) = (1\/2)3<\/sup> \u2013 6(1\/2)2<\/sup> + 2(1\/2) \u2013 4 = 1 + 1\/8 \u2013 4 \u2013 3\/2 = -35\/8<\/p>\n\n\n\n

Actual Division:<\/strong><\/p>\n\n\n\n

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

Question 6: f(x) = x4<\/sup> \u2013 3x2<\/sup> + 4, g(x) = x \u2013 2<\/strong><\/p>\n\n\n\n

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

f(x) = x4<\/sup> \u2013 3x2<\/sup> + 4, g(x) = x \u2013 2<\/p>\n\n\n\n

Put g(x) = 0<\/p>\n\n\n\n

\u21d2 x \u2013 2 = 0 or x = 2<\/p>\n\n\n\n

Remainder = f(2)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(2) = (2)4<\/sup> \u2013 3(2)2<\/sup> + 4 = 16 \u2013 12 + 4 = 8<\/p>\n\n\n\n

Actual Division:<\/strong><\/p>\n\n\n\n

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

Question 7: f(x) = 9x3<\/sup> \u2013 3x2<\/sup> + x \u2013 5, g(x) = x \u2013 2\/3<\/strong><\/p>\n\n\n\n

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

f(x) = 9x3<\/sup> \u2013 3x2<\/sup> + x \u2013 5, g(x) = x \u2013 2\/3<\/p>\n\n\n\n

Put g(x) = 0<\/p>\n\n\n\n

\u21d2 x \u2013 2\/3 = 0 or x = 2\/3<\/p>\n\n\n\n

Remainder = f(2\/3)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(2\/3) = 9(2\/3)3<\/sup> \u2013 3(2\/3)2<\/sup> + (2\/3) \u2013 5 = 8\/3 \u2013 4\/3 + 2\/3 \u2013 5\/1 = -3<\/p>\n\n\n\n

Actual Division:<\/strong><\/p>\n\n\n\n

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

Exercise 6.4 Page No: 6.24<\/h4>\n\n\n\n

In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7)<\/strong><\/p>\n\n\n\n

Question 1: f(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6; g(x) = x \u2013 3<\/strong><\/p>\n\n\n\n

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

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n\n\n\n

g(x) = x -3 = 0<\/p>\n\n\n\n

or x = 3<\/p>\n\n\n\n

Remainder = f(3)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(3) = (3)3<\/sup> \u2013 6(3)2<\/sup> +11 x 3 \u2013 6<\/p>\n\n\n\n

= 27 \u2013 54 + 33 \u2013 6<\/p>\n\n\n\n

= 60 \u2013 60<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, g(x) is a factor of f(x)<\/p>\n\n\n\n

Question 2: f(x) = 3X4<\/sup> + 17x3<\/sup> + 9x2<\/sup> \u2013 7x \u2013 10; g(x) = x + 5<\/strong><\/p>\n\n\n\n

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

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n\n\n\n

g(x) = x + 5 = 0, then x = -5<\/p>\n\n\n\n

Remainder = f(-5)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(3) = 3(-5)4<\/sup> + 17(-5)3<\/sup> + 9(-5)2<\/sup> \u2013 7(-5) \u2013 10<\/p>\n\n\n\n

= 3 x 625 + 17 x (-125) + 9 x (25) \u2013 7 x (-5) \u2013 10<\/p>\n\n\n\n

= 1875 -2125 + 225 + 35 \u2013 10<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, g(x) is a factor of f(x).<\/p>\n\n\n\n

Question 3: f(x) = x5<\/sup> + 3x4<\/sup> \u2013 x3<\/sup> \u2013 3x2<\/sup> + 5x + 15, g(x) = x + 3<\/strong><\/p>\n\n\n\n

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

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n\n\n\n

g(x) = x + 3 = 0, then x = -3<\/p>\n\n\n\n

Remainder = f(-3)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(-3) = (-3)5<\/sup> + 3(-3)4<\/sup> \u2013 (-3)3<\/sup> \u2013 3(-3)2<\/sup> + 5(-3) + 15<\/p>\n\n\n\n

= -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15<\/p>\n\n\n\n

= -243 +243 + 27-27- 15 + 15<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, g(x) is a factor of f(x).<\/p>\n\n\n\n

Question 4: f(x) = x3<\/sup> \u2013 6x2<\/sup> \u2013 19x + 84, g(x) = x \u2013 7<\/strong><\/p>\n\n\n\n

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

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n\n\n\n

g(x) = x \u2013 7 = 0, then x = 7<\/p>\n\n\n\n

Remainder = f(7)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(7) = (7)3<\/sup> \u2013 6(7)2<\/sup> \u2013 19 x 7 + 84<\/p>\n\n\n\n

= 343 \u2013 294 \u2013 133 + 84<\/p>\n\n\n\n

= 343 + 84 \u2013 294 \u2013 133<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, g(x) is a factor of f(x).<\/p>\n\n\n\n

Question 5: f(x) = 3x3<\/sup> + x2<\/sup> \u2013 20x + 12, g(x) = 3x \u2013 2<\/strong><\/p>\n\n\n\n

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

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n\n\n\n

g(x) = 3x \u2013 2 = 0, then x = 2\/3<\/p>\n\n\n\n

Remainder = f(2\/3)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(2\/3) = 3(2\/3) 3<\/sup> + (2\/3) 2<\/sup> \u2013 20(2\/3) + 12<\/p>\n\n\n\n

= 3 x 8\/27 + 4\/9 \u2013 40\/3 + 12<\/p>\n\n\n\n

= 8\/9 + 4\/9 \u2013 40\/3 + 12<\/p>\n\n\n\n

= 0\/9<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, g(x) is a factor of f(x).<\/p>\n\n\n\n

Question 6: f(x) = 2x3<\/sup> \u2013 9x2<\/sup> + x + 12, g(x) = 3 \u2013 2x<\/strong><\/p>\n\n\n\n

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

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n\n\n\n

g(x) = 3 \u2013 2x = 0, then x = 3\/2<\/p>\n\n\n\n

Remainder = f(3\/2)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(3\/2) = 2(3\/2)3<\/sup> \u2013 9(3\/2)2<\/sup> + (3\/2) + 12<\/p>\n\n\n\n

= 2 x 27\/8 \u2013 9 x 9\/4 + 3\/2 + 12<\/p>\n\n\n\n

= 27\/4 \u2013 81\/4 + 3\/2 + 12<\/p>\n\n\n\n

= 0\/4<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, g(x) is a factor of f(x).<\/p>\n\n\n\n

Question 7: f(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6, g(x) = x2<\/sup> \u2013 3x + 2<\/strong><\/p>\n\n\n\n

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

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n\n\n\n

g(x) = 0<\/p>\n\n\n\n

or x2<\/sup> \u2013 3x + 2 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 x \u2013 2x + 2 = 0<\/p>\n\n\n\n

x(x \u2013 1) \u2013 2(x \u2013 1) = 0<\/p>\n\n\n\n

(x \u2013 1) (x \u2013 2) = 0<\/p>\n\n\n\n

Therefore x = 1 or x = 2<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\u200df(1) = (1)3<\/sup> \u2013 6(1)2<\/sup> + 11(1) \u2013 6 = 1-6+11-6= 12- 12 = 0<\/p>\n\n\n\n

f(2) = (2)3<\/sup> \u2013 6(2)2<\/sup> + 11(2) \u2013 6 = 8 \u2013 24 + 22 \u2013 6 = 30 \u2013 30 = 0<\/p>\n\n\n\n

\u21d2 f(1) = 0 and f(2) = 0<\/p>\n\n\n\n

\u200dWhich implies g(x) is factor of f(x).<\/p>\n\n\n\n

Question 8: Show that (x \u2013 2), (x + 3) and (x \u2013 4) are factors of x3<\/sup> \u2013 3x2<\/sup> \u2013 10x + 24.<\/strong><\/p>\n\n\n\n

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

Let f(x) = x3<\/sup> \u2013 3x2<\/sup> \u2013 10x + 24<\/p>\n\n\n\n

If x \u2013 2 = 0, then x = 2,<\/p>\n\n\n\n

If x + 3 = 0 then x = -3,<\/p>\n\n\n\n

and If x \u2013 4 = 0 then x = 4<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(2) = (2)3<\/sup> \u2013 3(2)2<\/sup> \u2013 10 x 2 + 24 = 8 \u2013 12 \u2013 20 + 24 = 32 \u2013 32 = 0<\/p>\n\n\n\n

\u200d<\/p>\n\n\n\n

f(-3) = (-3)3<\/sup> \u2013 3(-3)2<\/sup> \u2013 10 (-3) + 24 = -27 -27 + 30 + 24 = -54 + 54 = 0<\/p>\n\n\n\n

f(4) = (4)3<\/sup> \u2013 3(4)2<\/sup> \u2013 10 x 4 + 24 = 64-48 -40 + 24 = 88 \u2013 88 = 0<\/p>\n\n\n\n

f(2) = 0<\/p>\n\n\n\n

f(-3) = 0<\/p>\n\n\n\n

f(4) = 0<\/p>\n\n\n\n

Hence (x \u2013 2), (x + 3) and (x \u2013 4) are the factors of f(x)<\/p>\n\n\n\n

Question 9: Show that (x + 4), (x \u2013 3) and (x \u2013 7) are factors of x3<\/sup> \u2013 6x2<\/sup> \u2013 19x + 84.<\/strong><\/p>\n\n\n\n

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

Let f(x) = x3<\/sup> \u2013 6x2<\/sup> \u2013 19x + 84<\/p>\n\n\n\n

If x + 4 = 0, then x = -4<\/p>\n\n\n\n

If x \u2013 3 = 0, then x = 3<\/p>\n\n\n\n

and if x \u2013 7 = 0, then x = 7<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(-4) = (-4)3<\/sup> \u2013 6(-4)2<\/sup> \u2013 19(-4) + 84 = -64 \u2013 96 + 76 + 84 = 160 \u2013 160 = 0<\/p>\n\n\n\n

f(-4) = 0<\/p>\n\n\n\n

f(3) = (3) 3<\/sup> \u2013 6(3) 2<\/sup> \u2013 19 x 3 + 84 = 27 \u2013 54 \u2013 57 + 84 = 111 -111=0<\/p>\n\n\n\n

f(3) = 0<\/p>\n\n\n\n

f(7) = (7) 3<\/sup> \u2013 6(7) 2<\/sup> \u2013 19 x 7 + 84 = 343 \u2013 294 \u2013 133 + 84 = 427 \u2013 427 = 0<\/p>\n\n\n\n

f(7) = 0<\/p>\n\n\n\n

Hence (x + 4), (x \u2013 3), (x \u2013 7) are the factors of f(x).<\/p>\n\n\n\n

\n\n\n\n

Exercise 6.5 Page No: 6.32<\/h4>\n\n\n\n

Using factor theorem, factorize each of the following polynomials:<\/strong><\/p>\n\n\n\n

Question 1: x3<\/sup> + 6x2<\/sup> + 11x + 6<\/strong><\/p>\n\n\n\n

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

Let f(x) = x3<\/sup> + 6x2<\/sup> + 11x + 6<\/p>\n\n\n\n

Step 1: Find the factors of constant term<\/p>\n\n\n\n

Here constant term = 6<\/p>\n\n\n\n

Factors of 6 are \u00b11, \u00b12, \u00b13, \u00b16<\/p>\n\n\n\n

Step 2: Find the factors of f(x)<\/p>\n\n\n\n

Let x + 1 = 0<\/p>\n\n\n\n

\u21d2 x = -1<\/p>\n\n\n\n

Put the value of x in f(x)<\/p>\n\n\n\n

f(-1) = (\u22121)3<\/sup> + 6(\u22121)2<\/sup> + 11(\u22121) + 6<\/p>\n\n\n\n

= -1 + 6 -11 + 6<\/p>\n\n\n\n

= 12 \u2013 12<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

So, (x + 1) is the factor of f(x)<\/p>\n\n\n\n

Let x + 2 = 0<\/p>\n\n\n\n

\u21d2 x = -2<\/p>\n\n\n\n

Put the value of x in f(x)<\/p>\n\n\n\n

f(-2) = (\u22122)3<\/sup> + 6(\u22122)2<\/sup> + 11(\u22122) + 6 = -8 + 24 \u2013 22 + 6 = 0<\/p>\n\n\n\n

So, (x + 2) is the factor of f(x)<\/p>\n\n\n\n

Let x + 3 = 0<\/p>\n\n\n\n

\u21d2 x = -3<\/p>\n\n\n\n

Put the value of x in f(x)<\/p>\n\n\n\n

f(-3) = (\u22123)3<\/sup> + 6(\u22123)2<\/sup> + 11(\u22123) + 6 = -27 + 54 \u2013 33 + 6 = 0<\/p>\n\n\n\n

So, (x + 3) is the factor of f(x)<\/p>\n\n\n\n

Hence, f(x) = (x + 1)(x + 2)(x + 3)<\/p>\n\n\n\n

Question 2: x3<\/sup> + 2x2<\/sup> \u2013 x \u2013 2<\/strong><\/p>\n\n\n\n

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

Let f(x) = x3<\/sup> + 2x2<\/sup> \u2013 x \u2013 2<\/p>\n\n\n\n

Constant term = -2<\/p>\n\n\n\n

Factors of -2 are \u00b11, \u00b12<\/p>\n\n\n\n

Let x \u2013 1 = 0<\/p>\n\n\n\n

\u21d2 x = 1<\/p>\n\n\n\n

Put the value of x in f(x)<\/p>\n\n\n\n

f(1) = (1)3<\/sup> + 2(1)2<\/sup> \u2013 1 \u2013 2 = 1 + 2 \u2013 1 \u2013 2 = 0<\/p>\n\n\n\n

So, (x \u2013 1) is factor of f(x)<\/p>\n\n\n\n

Let x + 1 = 0<\/p>\n\n\n\n

\u21d2 x = -1<\/p>\n\n\n\n

Put the value of x in f(x)<\/p>\n\n\n\n

f(-1) = (-1)3<\/sup> + 2(-1)2<\/sup> \u2013 1 \u2013 2 = -1 + 2 + 1 \u2013 2 = 0<\/p>\n\n\n\n

(x + 1) is a factor of f(x)<\/p>\n\n\n\n

Let x + 2 = 0<\/p>\n\n\n\n

\u21d2 x = -2<\/p>\n\n\n\n

Put the value of x in f(x)<\/p>\n\n\n\n

f(-2) = (-2)3<\/sup> + 2(-2)2<\/sup> \u2013 (-2) \u2013 2 = -8 + 8 + 2 \u2013 2 = 0<\/p>\n\n\n\n

(x + 2) is a factor of f(x)<\/p>\n\n\n\n

Let x \u2013 2 = 0<\/p>\n\n\n\n

\u21d2 x = 2<\/p>\n\n\n\n

Put the value of x in f(x)<\/p>\n\n\n\n

f(2) = (2)3<\/sup> + 2(2)2<\/sup> \u2013 2 \u2013 2 = 8 + 8 \u2013 2 \u2013 2 = 12 \u2260 0<\/p>\n\n\n\n

(x \u2013 2) is not a factor of f(x)<\/p>\n\n\n\n

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

Question 3: x3<\/sup> \u2013 6x2<\/sup> + 3x + 10<\/strong><\/p>\n\n\n\n

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

Let f(x) = x3<\/sup> \u2013 6x2<\/sup> + 3x + 10<\/p>\n\n\n\n

Constant term = 10<\/p>\n\n\n\n

Factors of 10 are \u00b11, \u00b12, \u00b15, \u00b110<\/p>\n\n\n\n

Let x + 1 = 0 or x = -1<\/p>\n\n\n\n

f(-1) = (-1)3<\/sup> \u2013 6(-1)2<\/sup> + 3(-1) + 10 = 10 \u2013 10 = 0<\/p>\n\n\n\n

f(-1) = 0<\/p>\n\n\n\n

Let x + 2 = 0 or x = -2<\/p>\n\n\n\n

f(-2) = (-2)3<\/sup> \u2013 6(-2)2<\/sup> + 3(-2) + 10 = -8 \u2013 24 \u2013 6 + 10 = -28<\/p>\n\n\n\n

f(-2) \u2260 0<\/p>\n\n\n\n

Let x \u2013 2 = 0 or x = 2<\/p>\n\n\n\n

f(2) = (2)3<\/sup> \u2013 6(2)2<\/sup> + 3(2) + 10 = 8 \u2013 24 + 6 + 10 = 0<\/p>\n\n\n\n

f(2) = 0<\/p>\n\n\n\n

Let x \u2013 5 = 0 or x = 5<\/p>\n\n\n\n

f(5) = (5)3<\/sup> \u2013 6(5)2<\/sup> + 3(5) + 10 = 125 \u2013 150 + 15 + 10 = 0<\/p>\n\n\n\n

f(5) = 0<\/p>\n\n\n\n

Therefore, (x + 1), (x \u2013 2) and (x-5) are factors of f(x)<\/p>\n\n\n\n

Hence f(x) = (x + 1) (x \u2013 2) (x-5)<\/p>\n\n\n\n

Question 4: x4<\/sup> \u2013 7x3<\/sup> + 9x2<\/sup> + 7x- 10<\/strong><\/p>\n\n\n\n

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

Let f(x) = x4<\/sup> \u2013 7x3<\/sup> + 9x2<\/sup> + 7x- 10<\/p>\n\n\n\n

Constant term = -10<\/p>\n\n\n\n

Factors of -10 are \u00b11, \u00b12, \u00b15, \u00b110<\/p>\n\n\n\n

Let x \u2013 1 = 0 or x = 1<\/p>\n\n\n\n

f(1) = (1)4<\/sup> \u2013 7(1)3<\/sup> + 9(1)2<\/sup> + 7(1) \u2013 10 = 1 \u2013 7 + 9 + 7 -10 = 0<\/p>\n\n\n\n

f(1) = 0<\/p>\n\n\n\n

Let x + 1 = 0 or x = -1<\/p>\n\n\n\n

f(-1) = (-1)4<\/sup> \u2013 7(-1)3<\/sup> + 9(-1)2<\/sup> + 7(-1) \u2013 10 = 1 + 7 + 9 \u2013 7 -10 = 0<\/p>\n\n\n\n

f(-1) = 0<\/p>\n\n\n\n

Let x \u2013 2 = 0 or x = 2<\/p>\n\n\n\n

f(2) = (2)4<\/sup> \u2013 7(2)3<\/sup> + 9(2)2<\/sup> + 7(2) \u2013 10 = 16 \u2013 56 + 36 + 14 \u2013 10 = 0<\/p>\n\n\n\n

f(2) = 0<\/p>\n\n\n\n

Let x \u2013 5 = 0 or x = 5<\/p>\n\n\n\n

f(5) = (5)4<\/sup> \u2013 7(5)3<\/sup> + 9(5)2<\/sup> + 7(5) \u2013 10 = 625 \u2013 875 + 225 + 35 \u2013 10 = 0<\/p>\n\n\n\n

f(5) = 0<\/p>\n\n\n\n

Therefore, (x \u2013 1), (x + 1), (x \u2013 2) and (x-5) are factors of f(x)<\/p>\n\n\n\n

Hence f(x) = (x \u2013 1) (x + 1) (x \u2013 2) (x-5)<\/p>\n\n\n\n

Question 5: x4<\/sup> \u2013 2x3<\/sup> \u2013 7x2<\/sup> + 8x + 12<\/strong><\/p>\n\n\n\n

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

f(x) = x4<\/sup> \u2013 2x3<\/sup> \u2013 7x2<\/sup> + 8x + 12<\/p>\n\n\n\n

Constant term = 12<\/p>\n\n\n\n

Factors of 12 are \u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b112<\/p>\n\n\n\n

Let x \u2013 1 = 0 or x = 1<\/p>\n\n\n\n

f(1) = (1)4<\/sup> \u2013 2(1)3<\/sup> \u2013 7(1)2<\/sup> + 8(1) + 12 = 1 \u2013 2 \u2013 7 + 8 + 12 = 12<\/p>\n\n\n\n

f(1) \u2260 0<\/p>\n\n\n\n

Let x + 1 = 0 or x = -1<\/p>\n\n\n\n

f(-1) = (-1)4<\/sup> \u2013 2(-1)3<\/sup> \u2013 7(-1)2<\/sup> + 8(-1) + 12 = 1 + 2 \u2013 7 \u2013 8 + 12 = 0<\/p>\n\n\n\n

f(-1) = 0<\/p>\n\n\n\n

Let x +2 = 0 or x = -2<\/p>\n\n\n\n

f(-2) = (-2)4<\/sup> \u2013 2(-2)3<\/sup> \u2013 7(-2)2<\/sup> + 8(-2) + 12 = 16 + 16 \u2013 28 \u2013 16 + 12 = 0<\/p>\n\n\n\n

f(-2) = 0<\/p>\n\n\n\n

Let x \u2013 2 = 0 or x = 2<\/p>\n\n\n\n

f(2) = (2)4<\/sup> \u2013 2(2)3<\/sup> \u2013 7(2)2<\/sup> + 8(2) + 12 = 16 \u2013 16 \u2013 28 + 16 + 12 = 0<\/p>\n\n\n\n

f(2) = 0<\/p>\n\n\n\n

Let x \u2013 3 = 0 or x = 3<\/p>\n\n\n\n

f(3) = (3)4<\/sup> \u2013 2(3)3<\/sup> \u2013 7(3)2<\/sup> + 8(3) + 12 = 0<\/p>\n\n\n\n

f(3) = 0<\/p>\n\n\n\n

Therefore, (x + 1), (x + 2), (x \u2013 2) and (x-3) are factors of f(x)<\/p>\n\n\n\n

Hence f(x) = (x + 1)(x + 2) (x \u2013 2) (x-3)<\/p>\n\n\n\n

Question 6: x4<\/sup> + 10x3<\/sup> + 35x2<\/sup> + 50x + 24<\/strong><\/p>\n\n\n\n

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

Let f(x) = x4<\/sup> + 10x3<\/sup> + 35x2<\/sup> + 50x + 24<\/p>\n\n\n\n

Constant term = 24<\/p>\n\n\n\n

Factors of 24 are \u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b18, \u00b112, \u00b124<\/p>\n\n\n\n

Let x + 1 = 0 or x = -1<\/p>\n\n\n\n

f(-1) = (-1)4<\/sup> + 10(-1)3<\/sup> + 35(-1)2<\/sup> + 50(-1) + 24 = 1 \u2013 10 + 35 \u2013 50 + 24 = 0<\/p>\n\n\n\n

f(1) = 0<\/p>\n\n\n\n

(x + 1) is a factor of f(x)<\/p>\n\n\n\n

Likewise, (x + 2),(x + 3),(x + 4) are also the factors of f(x)<\/p>\n\n\n\n

Hence f(x) = (x + 1) (x + 2)(x + 3)(x + 4)<\/p>\n\n\n\n

Question 7: 2x4<\/sup> \u2013 7x3<\/sup> \u2013 13x2<\/sup> + 63x \u2013 45<\/strong><\/p>\n\n\n\n

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

Let f(x) = 2x4<\/sup> \u2013 7x3<\/sup> \u2013 13x2<\/sup> + 63x \u2013 45<\/p>\n\n\n\n

Constant term = -45<\/p>\n\n\n\n

Factors of -45 are \u00b11, \u00b13, \u00b15, \u00b19, \u00b115, \u00b145<\/p>\n\n\n\n

Here coefficient of x^4 is 2. So possible rational roots of f(x) are<\/p>\n\n\n\n

\u00b11, \u00b13, \u00b15, \u00b19, \u00b115, \u00b145, \u00b11\/2,\u00b13\/2,\u00b15\/2,\u00b19\/2,\u00b115\/2,\u00b145\/2<\/p>\n\n\n\n

Let x \u2013 1 = 0 or x = 1<\/p>\n\n\n\n

f(1) = 2(1)4<\/sup> \u2013 7(1)3<\/sup> \u2013 13(1)2<\/sup> + 63(1) \u2013 45 = 2 \u2013 7 \u2013 13 + 63 \u2013 45 = 0<\/p>\n\n\n\n

f(1) = 0<\/p>\n\n\n\n

f(x) can be written as,<\/p>\n\n\n\n

f(x) = (x-1) (2x3<\/sup> \u2013 5x2<\/sup> -18x +45)<\/p>\n\n\n\n

or f(x) =(x-1)g(x) \u2026(1)<\/p>\n\n\n\n

Let x \u2013 3 = 0 or x = 3<\/p>\n\n\n\n

f(3) = 2(3)4<\/sup> \u2013 7(3)3<\/sup> \u2013 13(3)2<\/sup> + 63(3) \u2013 45 = = 162 \u2013 189 \u2013 117 + 189 \u2013 45= 0<\/p>\n\n\n\n

f(3) = 0<\/p>\n\n\n\n

Now, we are available with 2 factors of f(x), (x \u2013 1) and (x \u2013 3)<\/p>\n\n\n\n

Here g(x) = 2x2<\/sup> (x-3) + x(x-3) -15(x-3)<\/p>\n\n\n\n

Taking (x-3) as common<\/p>\n\n\n\n

= (x-3)(2x2<\/sup> + x \u2013 15)<\/p>\n\n\n\n

= (x-3)(2x2<\/sup>+6x \u2013 5x -15)<\/p>\n\n\n\n

= (x-3)(2x-5)(x+3)<\/p>\n\n\n\n

= (x-3)(x+3)(2x-5) \u2026.(2)<\/p>\n\n\n\n

From (1) and (2)<\/p>\n\n\n\n

f(x) =(x-1) (x-3)(x+3)(2x-5)<\/p>\n\n\n\n

\n\n\n\n

Exercise VSAQs Page No: 6.33<\/h4>\n\n\n\n

Question 1: Define zero or root of a polynomial<\/strong><\/p>\n\n\n\n

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

zero or root, is a solution to the polynomial equation, f(y) = 0.<\/p>\n\n\n\n

It is that value of y that makes the polynomial equal to zero.<\/p>\n\n\n\n

Question 2: If x = 1\/2 is a zero of the polynomial f(x) = 8x3<\/sup> + ax2<\/sup> \u2013 4x + 2, find the value of a.<\/strong><\/p>\n\n\n\n

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

If x = 1\/2 is a zero of the polynomial f(x), then f(1\/2) = 0<\/p>\n\n\n\n

8(1\/2)3<\/sup> + a(1\/2)2<\/sup> \u2013 4(1\/2) + 2 = 0<\/p>\n\n\n\n

8 x 1\/8 + a\/4 \u2013 2 + 2 = 0<\/p>\n\n\n\n

1 + a\/4 = 0<\/p>\n\n\n\n

a = -4<\/p>\n\n\n\n

Question 3: Write the remainder when the polynomial f(x) = x3<\/sup> + x2<\/sup> \u2013 3x + 2 is divided by x + 1.<\/strong><\/p>\n\n\n\n

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

Using factor theorem,<\/p>\n\n\n\n

Put x + 1 = 0 or x = -1<\/p>\n\n\n\n

f(-1) is the remainder.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(-1) = (-1)3<\/sup> + (-1)2<\/sup> \u2013 3(-1) + 2<\/p>\n\n\n\n

= -1 + 1 + 3 + 2<\/p>\n\n\n\n

= 5<\/p>\n\n\n\n

Therefore 5 is the remainder.<\/p>\n\n\n\n

Question 4: Find the remainder when x3<\/sup> + 4x2<\/sup> + 4x-3 if divided by x<\/strong><\/p>\n\n\n\n

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

Using factor theorem,<\/p>\n\n\n\n

Put x = 0<\/p>\n\n\n\n

f(0) is the remainder.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

f(0) = 03<\/sup> + 4(0)2<\/sup> + 4\u00d70 -3 = -3<\/p>\n\n\n\n

Therefore -3 is the remainder.<\/p>\n\n\n\n

Question 5: <\/strong>If x+1 is a factor of x3<\/sup> + a, then write the value of a.<\/p>\n\n\n\n

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

Let f(x) = x3<\/sup> + a<\/p>\n\n\n\n

If x+1 is a factor of x3<\/sup> + a then f(-1) = 0<\/p>\n\n\n\n

(-1)3<\/sup> + a = 0<\/p>\n\n\n\n

-1 + a = 0<\/p>\n\n\n\n

or a = 1<\/p>\n\n\n\n

Question 6: If f(x) = x^4 \u2013 2x3<\/sup> + 3x2<\/sup> \u2013 ax \u2013 b when divided by x \u2013 1, the remainder is 6, then find the value of a+b.<\/strong><\/p>\n\n\n\n


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

From the statement, we have f(1) = 6<\/p>\n\n\n\n

(1)^4 \u2013 2(1)3<\/sup> + 3(1)2<\/sup> \u2013 a(1) \u2013 b = 6<\/p>\n\n\n\n

1 \u2013 2 + 3 \u2013 a \u2013 b = 6<\/p>\n\n\n\n

2 \u2013 a \u2013 b = 6<\/p>\n\n\n\n

a + b = -4<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 6: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 6\u2013Factorization Of Polynomials<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 6\u2013Factorization Of Polynomials PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n