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