**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*) = 2*x*^{3} + *x*^{2} - 2*x* - 1, *g*(*x*) = *x* + 1

(ii) *p*(*x*) = *x*^{3} + 3*x*^{2} + 3*x* + 1, *g*(*x*) = *x* + 2

(iii) *p*(*x*) = *x*^{3} - 4 *x*^{2} + *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*) = 2*x*^{3} + *x*^{2} - 2*x* - 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 2*x*^{3} + *x*^{2} - 2*x* - 1.

(ii) *p*(*x*) = *x*^{3} +3*x*^{2} + 3*x* + 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 *x*^{3} +3*x*^{2} + 3*x* + 1.

(iii) *p*(*x*) = *x*^{3} - 4*x*^{2} + 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 *x*^{3} - 4*x*^{2} + *x* + 6.