Question

Find the remainder when $x^3+3x^2+3x+1$ is divided by
(i) $x+1$
(ii) $x-\frac{1}{2}$
(iii) $x$
(iv) $x+\pi$
(v) $5+2x$

Answer 1

Using remainder theorem, if $p\left(x\right)$ is divided by $(x-a)$, then the remainder can be found by $p\left(a\right)$

Given: $p\left(x\right)=x^3+3x^2+3x+1$.

(i) If $x+1=0$, then $x=-1$

So putting the value of $x=-1$ in the given equation, we get,

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

Hence, remainder is $0$.

(ii) If $x-\frac{1}{2}=0$, then $x=\frac{1}{2}$

So, putting $x=\frac{1}{2}$ in equation, we get

$p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^3+3{\left(\frac{1}{2}\right)}^2+3\left(\frac{1}{2}\right)+1=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1=\frac{27}{8}=3\frac{3}{8}$

Hence, the remainder is $3\frac{3}{8}$.

(iii) If $x+0=0$ then $x=0$

So putting the value of $x=0$ in the given equation, we get,

$p\left(0\right)=(0{)}^3+3(0{)}^2+3\left(0\right)+1=0+0+0+1=1$

Hence, the remainder is $1$.

(iv) If $x+\pi =0$, then $x=-\pi$

So putting the value of $x=-\pi$ in the equation ,we get,

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

Hence, the remainder is $-{\pi }^3+3{\pi }^2-3\pi +1$

(v) If $5+2x=0$, then $x=-\frac{2}{5}$

So putting the value of $x=-\frac{2}{5}$ in the given equation, we get,

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

$=-\frac{8}{125}+\frac{12}{25}-\frac{6}{5}+1=\frac{-8+60-150+125}{125}=\frac{27}{125}$

Hence, the remainder is $\frac{27}{125}$