Question

Given that $6x^2+5ax-12a$ leaves a remainder of - 4 when divided by (x - a), the possible value of a is ..........

• A. - 2
• B. $\frac{1}{2}$
• C. $-\frac{2}{3}$
• D. Both (A) and (C)

Answer 1

The correct option is D Both (A) and (C)

$6x^3-5ax+12a$ divided by (x-a) so put value x=a
then $6a^3-5a\times a+12a=0\Rightarrow 6a^3-5a^2-12a-4=0$
$\Rightarrow 6a^3-5a^2-12a-4=0\Rightarrow 6a^3-12a^2+7a^2-14a-2a-4\Rightarrow 6a^2(a-2)+7(a-2)+2(a-2)0$
$\Rightarrow (6a^2+7a+2)(a-2)=0\Rightarrow (a-2)(2a+1)(3a+2)=0$
Then a=2 ,a=-$\frac{1}{2}$ or a=-$\frac{2}{3}$