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

We will find the remainder by using the Remainder theorem.

Remainder theorem:

when a polynomial $p\left(x\right)$ is divided by the other polynomial $(x-a)$ then remainder is $p\left(a\right)$.

(i)

Let us denote the given polynomials as

$f\left(x\right)=x^3+3x^2+3x+1$

$g\left(x\right)=x+1$

$\Rightarrow g\left(x\right)=x-(-1)$

Remainder $=f(-1)$

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

$=-1+3-3+1$

$=0$

$\therefore f(-1)=0$

(ii)

Let us denote the given polynomials as

$f\left(x\right)=x^3+3x^2+3x+1$

$g\left(x\right)=x-\frac{1}{2}$

Remainder $=f\left(\frac{1}{2}\right)$

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

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

$=\frac{1+6+9+8}{8}$

$=\frac{24}{8}$

$=3$

$\therefore f\left(\frac{1}{2}\right)=3$

(iii)

Let us denote the given polynomials as,

$f\left(x\right)=x^3+3x^2+3x+1$

$g\left(x\right)=x$

$\Rightarrow g\left(x\right)=x-0$

Remainder $=f\left(0\right)$

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

$=1$

$\therefore f\left(0\right)=1$

(iv)

Let us denote the given polynomials as

$f\left(x\right)=x^3+3x^2+3x+1$

$g\left(x\right)=x+\pi$

$\Rightarrow g\left(x\right)=x-(-\pi )$

Remainder $=f(-\pi )$

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

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

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

(v)

Let us denote the given polynomials as

$f\left(x\right)=x^3+3x^2+3x+1$

$g\left(x\right)=5+2x$

Put $5+2x=0$

$\Rightarrow 2x=-5$

$\Rightarrow x=-\frac{5}{2}$

i.e., Remainder is $f(-\frac{5}{2})$

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

$=-\frac{125}{8}+3\left(\frac{25}{4}\right)-3\left(\frac{5}{2}\right)+1$

$=-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1$

$=\frac{-125+150-60+8}{8}$

$=-\frac{27}{8}$

$\therefore f(-\frac{5}{2})=-\frac{27}{8}$