NCERT Solutions
Class 9 Maths
Polynomials

Ex.2.4 Q.2
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(1) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
(2) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(3) p(x) = x3 - 4x2 + x + 6, g(x) = x - 3
(1) Given, p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
Put x + 1 = 0, we get x = -1
Using remainder theorem, when p(x) = 2x3 + x2 – 2x – 1 is divided by g(x) = x + 1, remainder is given by p (-1).
Now, p (-1) = 2 × (-1)3 + (-1)2 – 2 × (-1) – 1
= -2 + 1 + 2 – 1
= -3 + 3
= 0
Since remainder p (-1) = 0, hence g(x) is a factor of p(x).
(2) Given, p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Put x + 2 = 0, we get x = -2
Using remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by g(x) = x + 2, remainder is given by p (-2).
Now, p (-2) = (-2)3 + 3 × (-2)2 + 3 × (-2) + 1
= -8 + 3 × 4 – 3 × 2 + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
Since remainder p (-2) ≠ 0, hence g(x) is not a factor of p(x).
(3) Given, p(x) = x3 - 4x2 + x + 6, g(x) = x – 3
Put x - 3 = 0, we get x = 3
Using remainder theorem, when p(x) = x3 - 4x2 + x + 6 is divided by g(x) = x - 3, remainder is given by p (3).
Now, p (3) = 33 - 4 × 32 + 3 + 6
= 27 - 4 × 9 + 3 + 6
= 27 - 36 + 3 + 6
= 36 - 36
= 0
Since remainder p (3) = 0, hence g(x) is a factor of p(x).