Question

Find the remainder when the polynomial $x^3+3x^2+3x+1$ is divided by
$(x-\frac{1}{2})$.

Answer 1

Given, polynomial $p\left(x\right)=x^3+3x^2+3x+1$ and linear factor is $(x-\frac{1}{2})$
We find remainder using the remainder theorem that state that the remainder obtained by the polynomial $f\left(x\right)$ divided by the linear factor $(x-a)$ is $f\left(a\right)$
Compare $(x-a)$ with $(x-\frac{1}{2})$, we get $a=\frac{1}{2}$
Thus, the remainder is $=p\left(a\right)$
$=(\frac{1}{2}{)}^3+3(\frac{1}{2}{)}^2+3\left(\frac{1}{2}\right)+1$
$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$
$=\frac{1+6+12+8}{8}$
$=\frac{27}{8}$
Therefore, the remainder is $\frac{27}{8}$