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

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To do:

We have to 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) $5+2 x$

Solution:

The remainder theorem states that when a polynomial $p(x)$ is divided by a linear polynomial, $x - a$ the remainder of that division will be equivalent to $p(a)$.

Let $f(x) =x^{3}+3 x^{2}+3 x+1$

Therefore,

(i) Let $p(x) = x +1$

$=x-(-1)$

So, the remainder will be $f(-1)$.

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

$= -1+3(1)-3+1$

$=-1+1+3-3$

$=0$

Therefore, the remainder is $0$.

(ii) Let $q(x)=x-\frac{1}{2}$

So, the remainder will be $f(\frac{1}{2})$.

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

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

$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$

$=\frac{1+3\times2+3\times4+1\times8}{8}$       (LCM of $8,4,2$ and $1$ is $8$)

$=\frac{1+6+12+8}{8}$

$=\frac{27}{8}$

Therefore, the remainder is $\frac{27}{8}$.

(iii) Let $r(x) = x$

$=x-0$

So, the remainder will be $f(0)$.

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

$= 0+3(0)+0+1$

$=1$

Therefore, the remainder is $1$.

(iv) Let $s(x) = x+\pi$

$=x-(-\pi)$

So, the remainder will be $f(-\pi)$.

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

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

Therefore, the remainder is $-\pi^3+3\pi^2-3\pi+1$.

(v) Let $t(x) = 5+2x$

$2x+5=0$

$2x=-5$

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

So, the remainder will be $f(-\frac{5}{2})$.

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

$= -\frac{125}{8}+3(\frac{25}{4})-\frac{15}{2}+1$

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

$=\frac{-125+75\times2-15\times4+1\times8}{8}$         (LCM of $8,4,2$ and $1$ is $8$)

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

$=\frac{158-185}{8}$

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

Therefore, the remainder is $-\frac{27}{8}$.

Updated on 10-Oct-2022 13:39:07