## RD Sharma class 9 maths Solutions Factorization of Polynomials Exercise-6.2

- Factorization of Polynomials Chapter 6 Ex 6.1
- Factorization of Polynomials Chapter 6 Ex 6.2
- Factorization of Polynomials Chapter 6 Ex 6.3
- Factorization of Polynomials Chapter 6 Ex 6.4
- Factorization of Polynomials Chapter 6 Ex 6.5

### RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

Question 1.

If f(x) = 2x^{3} – 13x^{2} + 17x + 12, find

(i) f (2)

(ii) f (-3)

(iii) f(0)

Solution:

f(x) = 2x^{3} – 13x^{2} + 17x + 12

(i) f(2) = 2(2)^{3} – 13(2)^{2} + 17(2) + 12

= 2 x 8-13 x 4+17 x 2+12

= 16-52 + 34 + 12

= 62 – 52

= 10

(ii) f(-3) = 2(-3)^{3} – 13(-3)^{2} + 17 x (-3) + 12

= 2 x (-27) – 13 x 9 + 17 x (-3) + 12

= -54 – 117 -51 + 12

= -222 + 12

= -210

(iii) f(0) = 2 x (0)^{3} – 13(0)^{2} + 17 x 0 + 12

= 0-0 + 0+ 12 = 12

Question 2.

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases: [NCERT]

Solution:

Question 3.

If x = 2 is a root of the polynomial f(x) = 2x^{2}-3x + la, find the value of a.

Solution:

p(x) = 2x^{2} – 3x + 7a

∵ x = 2 is its zero, then

p(0) = 0

∴ p( 2) = 2(2)^{2} – 3×2 + la = 0

⇒2 x 4-3 x2 + 7a = 0

⇒ 8 – 6 + 7o = 0

⇒2 + 7a = 0

⇒ 7a = -2 ⇒ a =\(\frac { -2 }{ 7 }\)

∴ Hence a = \(\frac { -2 }{ 7 }\)

Question 4.

If x = –\(\frac { 1 }{ 2 }\) is a zero of the polynomial p(x) = 8x^{3} – ax^{2} – x + 2, find the value of a.

Solution:

Question 5.

If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x^{3} – 3x^{2} + ax + b, find the value of a and b.

Solution:

f(x) = 2x^{3} – 3x^{2} + ax + b

∵ x = 0 and x = -1 are its zeros

∴ f(0) = 0 and f(-1) = 0

Now, f(0) = 0

⇒ 2(0)^{3} – 3(0)^{2} + a x 0 + b = 0

⇒ 0-0 + 0 + b= 0

∴ b = 0

and f(-1) = 0

⇒ 2(-1)^{3} – 3(-1)^{2} + a(-1) + b = 0

⇒ 2 x (-1) – 3 x 1 + a x (-1) + b = 0

⇒ -2 -3-a + b = 0

⇒ -2-3-a + 0 = 0

⇒ -5- a = 0=>a =-5

Hence a = -5, b = 0

Question 6.

Find the integral roots of the polynomial f(x) = x^{3} + 6x^{2} + 11x + 6.

Solution:

f(x) = x^{3} + 6x^{2} + 11x + 6

Construct = 6 = ±1, ±2, +3, ±6

If x = 1, then

f(1) = (1)^{3} + 6(1)^{2} + 11 x 1 + 6

= 1+ 6+11+ 6 = 24

∵ f(x) ≠ 0, +0

∴ x = 1 is not its zero

Similarly, f(-1) = (-1)^{3} + 6(-1)^{2} + 11(-1) + 6

= -1 + 6 x 1-11+6

=-1+6-11+6

= 12-12 = 0

∴ x = -1 is its zero

f(-2) = (-2)^{3} + 6(-2)^{2} + 11 (-2) + 6

= -8 + 24 – 22 + 6

= -30 + 30 = 0

∴ x = -2 is its zero

f(-3) = (-3)^{3} + 6(-3)^{2} + 11 (-3) + 6

= -27 + 54 – 33 + 6 = 60 – 60 = 0

∴ x = -3 is its zero

x = -1, -2, -3 are zeros of f(x)

Hence roots of f(x) are -1, -2, -3

Question 7.

Find the rational roots of the polynomial f(x) = 2x^{3} + x^{2} – 7x – 6.

Solution: