Question

Perform the following divisions and give the remainders.
$(3p^2+6p+18)÷(p-6)$

Answer 1

According to question,

$\frac{(3p^2+6p+18)}{p-6}$

$⟹\frac{3\left[\right(p-6\left)\right(p+8)+54]}{p-6}$

$⟹3(p+8)+\frac{162}{(p-6)}$

Here, Numerator is not completely divisible by denominator.

Hence, remainder is $162$