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

In this chapter, we provide RD Sharma Solutions for Chapter 6 Factorisation of Polynomials Ex 6.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 6 Factorisation of Polynomials Ex 6.2 Maths pdf, free RD Sharma Solutions for Chapter 6 Factorisation of Polynomials Ex 6.2 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Solutions for Class 9 Chapter 6 Factorisation of Polynomials Ex 6.2 Download PDF

Question 1.
If f(x) = 2x³ – 13x² + 17x + 12, find
(i) f (2)
(ii) f (-3)
(iii) f(0)
Solution:
f(x) = 2x³ – 13x² + 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) = 2x²-3x + la, find the value of a.
Solution:
p(x) = 2x² – 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 =−27
∴ Hence a = −27

Question 4.
If x = –12 is a zero of the polynomial p(x) = 8x³ – ax² – x + 2, find the value of a.
Solution:

Question 5.
If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x³ – 3x² + ax + b, find the value of a and b.
Solution:
f(x) = 2x³ – 3x² + 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) = x³ + 6x² + 11x + 6.
Solution:
f(x) = x³ + 6x² + 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) = 2x³ + x² – 7x – 6.
Solution:

All Chapter RD Sharma Solutions For Class 9 Maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 9

All Subject NCERT Solutions For Class 9

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.