Question

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

Answer 1

$p\left(x\right)=x^3+3x^2+3x+1$
(i) When p(x) is divided by x + 1, the remainder is p(-1),
$p(-1)=(-1{)}^3+3(-1{)}^2+3(-1)+1=-1+3-3+1=0$
Remainder = 0

(ii) When p(x) is divided by $x-\frac{1}{2}$, the remainder is $p\left(\frac{1}{2}\right).p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^3+3\left(\frac{1}{4}\right)+3\left(\frac{1}{2}\right)+1$
$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1=\frac{1+6+12+8}{8}$
$\therefore$ Remainder $=\frac{27}{8}=3\frac{3}{8}$