Question

What must be added to $x^3-3x^2-12x+19$ so that the result is exactly divisible by $x^2+x-6$?

• A. $2x-5$
• B. $2x+5$
• C. $-2x-5$
• D. $x+5$

Answer 1

Answer: (B)

$2x+5$

Given: $f\left(x\right)=x^3-3x^2-12x+19$
$g\left(x\right)=x^2+x-6$
By division algorithm
$f\left(x\right)=g\left(x\right)\times q\left(x\right)+r\left(x\right)$
$q\left(x\right)=\frac{f\left(x\right)+[-r(x\left)\right]}{g\left(x\right)}$
If means $r\left(x\right)$ must be added so that result is exactly divisible by $x^2+x-6$.
So, $2x+5$ must be added to $f\left(x\right)$, so that the result is exactly divisible by $x^2+x-6$.