Question

When $4x^3+2x^2-px+5$ is divided by $2x+6$, the remainder is $2$; and $3x+12$ is the factor of $q{x}^2-16$. Find $p-30q$.

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

The remainder when $4x^3+2x^2-px+5$ is divided by $2x+6$, is $2$.

$2x+6=0⟹2x=-6$
$⟹x=-3$
Substituting $x=-3$ in $4x^3+2x^2-px+5$, we should get $2$.
$⟹4(-3{)}^3+2(-3{)}^2-(-3)p+5=2$
$⟹-108+18+3p+5=2$
$⟹-85+3p=2$
$⟹3p=87$
$⟹p=\frac{87}{3}=29$

Also, $3x+12$ is a facto-r of $q{x}^2-16$
$3x+12=0⟹x=\frac{-12}{3}=-4$
Substituting $x=-4$ in $q{x}^2-16$, we should get $0$.
$⟹q(-4{)}^2-16=0$
$⟹16q=16$
$⟹q=\frac{16}{16}=1$

$p-30q=29-30\left(1\right)=29-30=-1$