Question

Given that $x^2+x-6$ is a factor of $2x^4+x^3-a{x}^2+bx+a+b-1,$ then which of the following is/are true?

• A. $b=3$
• B. $a=3$
• C. $a=16$
• D. $b=16$

Answer 1

Answer: (A), (C)

$a=16$

Let $f\left(x\right)=2x^4+x^3-a{x}^2+bx+a+b-1$
and $g\left(x\right)=x^2+x-6$

Taking $g\left(x\right)=x^2+x-6=(x+3)(x-2)$
$\therefore g\left(2\right)=g(-3)=0$

It is given that $g\left(x\right)$ is a factor of $f\left(x\right)$
So $f(-3)=f\left(2\right)=0$
$\Rightarrow f(-3)=2(-3{)}^4+(-3{)}^3-a(-3{)}^2+b(-3)+a+b-1=0$
$\Rightarrow 134-8a-2b=0$
$\Rightarrow 4a+b=\mathrm{67......}\left(i\right)$

$f\left(2\right)=2(2{)}^4+(2{)}^3-a(2{)}^2+b(2)+a+b-1=0$
$\Rightarrow 39-3a+3b=0$
$\Rightarrow a-b=\mathrm{13......}\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$a=16,b=3$