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

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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: