Question

When $x^3+5x^2+ax-7$ is divided by $x-3$, the remainder is $47$. Find $a$.

• A. $-6$
• B. $-2$
• C. $-3$
• D. $-4$

Answer 1

The correct option is A $-6$

$p\left(x\right)=x^3+5x^2+ax-7$
Divisor is $x-3$
$\therefore$ put $x=3$ in p(x)
Remainder = $p\left(3\right)=3^3+5(3{)}^2+a(3)-7$
= $27+5\left(9\right)+3a-7$
= $27+45+3a-7$
= $65+3a$
But remainder = 47
$\therefore 65+3a=47$ ....$\left[p\right(3)=47]$
$3a=47-65$
$3a=-18$
$\therefore a=-6$