Question

Find the remainder when $x^3+{3x}^2+3x+1$ is divided by:
$\left(i\right)x+1$
$\left(ii\right)5+2x$ [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

(i) Let $p\left(x\right)=x^3+{3x}^2+3x+1$ and $q\left(x\right)=x+1$

According to Remainder Theorem, remainder is equal to p(a) when p(x) is divided by (x−a).

$p(-1)={(-1)}^3+3{(-1)}^2+3(-1)+1=-1+3-3+1=0$

Therefore, remainder is 0 when $p\left(x\right)=x^3+{3x}^2+3x+1$ is divided by $x+1$

(ii)Let $p\left(x\right)=x^3+{3x}^2+3x+1$ and $q\left(x\right)=5+2x$

$p\left(\frac{-5}{2}\right)=(\frac{-5}{2}{)}^3+3(\frac{-5}{2}{)}^2+3\left(\frac{-5}{2}\right)+1$

$=\left(\frac{-125}{8}\right)+\left(\frac{75}{4}\right)+\left(\frac{-15}{2}\right)+1$

$=\left(\frac{-125+150-60+8}{8}\right)=\left(\frac{-27}{8}\right)$

Therefore, remainder is equal to $\left(\frac{-27}{8}\right)$ when $p\left(x\right)=x^3+{3x}^2+3x+1$ is divided by $5+2x$ (Using Remainder Theorem)