Q.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:

If g(x) = (x - a) is a factor of p(x), p(a) must be zero.

(i) p(x) = 2x3 + x2 - 2x - 1, g(x) = x + 1

Put g(x) = x + 1 = 0 ⇒ x = -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 2x3 + x2 - 2x - 1.

(ii) p(x) = x3 +3x2 + 3x + 1, g(x) = x + 2=

Put g(x) = x + 2= 0 ⇒ x = -2

p(-2) = (-2)3 + 3(- 2)2 + 3(- 2) + 1

= -8 + 12 - 6 + 1

= -1 ≠ 0

Hence g(x) = x + 2 is not a factor of x3 +3x2 + 3x + 1.

(iii) p(x) = x3 - 4x2 + x + 6, g(x) = x - 3

Put g(x) = x - 3 = 0 ⇒ x = 3

p(3) = (3)3 - 4(3)2 + 3 + 6

= 27 - 36 + 9 = 0

Therefore, g(x) = x - 3 is a factor of x3 - 4x2 + x + 6.