Question

The remainder when $f\left(x\right)=x^2+6x+8$ is divided by $2x+1$ is $\frac{p}{q}$, where $p$ and $q$ are co-primes. Find the value of $p-q$.

Answer 1

Given polynomial is $f\left(x\right)=x^2+6x+8$ and divisor is $2x+1$.
$\therefore$ Remainder = $f(-\frac{1}{2})={(-\frac{1}{2})}^2+6(-\frac{1}{2})+8$
$=\frac{1}{4}-3+8=5\frac{1}{4}=\frac{21}{4}=\frac{p}{q}$

$p-q=21-4=17$