# Polynomials - CBSE Guide NCERT Solutions of Class 9 Mathematics Exercise 2.4

## CBSE Guide NCERT Solutions of Class 9 Mathematics

### Chapter 2, POLYNOMIALS

#### NCERT Solutions of Math Exercise 2.4

Question.1: Determine which of the following polynomials has (x + 1) a factor:

Solution:
(i) To have (x + 1) a factor of p(x) = x3 + x2 + x + 1, substituting x = –1 must give p(−1) = 0, otherwise (x + 1) is not a factor of p(x).
p(x) = x3 + x2 + x + 1
p(−1) = (−1)3 + (−1)2 + (−1) + 1
= − 1 + 1 − 1 − 1 = 0
Therefore, x + 1 is a factor of this polynomial.
(ii) For (x + 1) to be a factor of p(x) = x4 + x3 + x2 + x + 1, p (−1) must be zero. Substituting x = –1 in the given polynomial we get,
p(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1
= 1 − 1 + 1 −1 + 1 = 1
As p(− 1) ≠ 0, (x + 1) is not a factor of this polynomial.

(iii)Taking cue from the above try yourself.

Question.2: Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x3 + x2 − 2x − 1, g(x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) p(x) = x3 − 4 x2 + x + 6, g(x) = x − 3
Solution:
(i) To have g(x) = x + 1 as a factor of the given polynomial p(x), p(−1) must be zero. ( x = –1)
Substituting x = –1 in 2x3 + x2 − 2x − 1
p(−1) = 2(−1)3 + (−1)2 − 2(−1) − 1
= 2(−1) + 1 + 2 − 1 = 0
Hence, g(x) = x + 1 is a factor of the given polynomial.
(ii) p(x) = x3 +3x2 + 3x + 1
g(x) = x + 2
x = –2, substituting this in the given polynomial p(x),
p(−2) = (−2)3 + 3(−2)2 + 3(−2) + 1
= − 8 + 12 − 6 + 1
= −1
Since p(−2) ≠ 0,therefore, g(x) = x + 2 is not a factor of the given polynomial.
(iii) (Taking hint from the above try yourself).

Question.3: Find the value of k, if x − 1 is a factor of p(x) in each of the following cases:

Solution:
As it is given that (x – 1) is a factor of p(x), so after substituting x = 1 in p(x) we can solve for k by making p(1) = 0 in each case.

Question.4: Factorize
(i) 12x2 − 7x + 1    (ii) 2x2 + 7x + 3
(iii) 6x2 + 5x − 6  (iv) 3x2 − x − 4
Solution:

Question.5: Factorize
(i) x3 − 2x2 − x + 2     (ii) x3 + 3x2 −9x − 5
(iii) x3 + 13x2 + 32x + 20  (iv) 2y3 + y2 − 2y − 1
Solution: