 QUESTION

# Find the remainder when the polynomial ${x^3} + 3{x^2} + 3x + 1$ is divided by each of the following:$(i) x + 1$ $(ii) x - \dfrac{1}{2}$ $(iii) x$ $(iv) x + \pi$ $(v) 5 + 2x$

Hint: If any polynomial is divided by $x + a,$then we can find the remainder by putting $x + a = 0$ i.e. $x = - a$ in the given polynomial.

The given polynomial is ${x^3} + 3{x^2} + 3x + 1$.
$(i)$When the polynomial is divided by $x + 1$, we can find the remainder by putting $x + 1 = 0$ i.e. $x = - 1$ in the polynomial:
$\Rightarrow {\text{Remainder}} = {( - 1)^3} + 3{( - 1)^2} + 3( - 1) + 1, \\ \Rightarrow {\text{Remainder}} = - 1 + 3 - 3 + 1, \\ \Rightarrow {\text{Remainder}} = 0 \\$
Thus, the remainder in this case is $0.$
$(ii)$ When the polynomial is divided by $x - \dfrac{1}{2}$, we can find the remainder by putting $x = \dfrac{1}{2}$ in the polynomial:
$\Rightarrow {\text{Remainder}} = {(\dfrac{1}{2})^3} + 3{(\dfrac{1}{2})^2} + 3(\dfrac{1}{2}) + 1, \\ \Rightarrow {\text{Remainder}} = \dfrac{1}{8} + \dfrac{3}{4} + \dfrac{3}{2} + 1, \\ \Rightarrow {\text{Remainder}} = \dfrac{{27}}{8}. \\$
Thus, the remainder in this case is $\dfrac{{27}}{8}.$
$(iii)$ When the polynomial is divided by $x$, we can find the remainder by putting $x = 0$ in the polynomial:
$\Rightarrow {\text{Remainder}} = {(0)^3} + 3{(0)^2} + 3(0) + 1, \\ \Rightarrow {\text{Remainder}} = 1. \\$
Thus, the remainder in this case is $1.$
$(iv)$ When the polynomial is divided by $x + \pi$, we can find the remainder by putting $x = - \pi$ in the polynomial:
$\Rightarrow {\text{Remainder}} = {( - \pi )^3} + 3{( - \pi )^2} + 3( - \pi ) + 1, \\ \Rightarrow {\text{Remainder}} = - {\pi ^3} + 3{\pi ^2} - 3\pi + 1. \\$
Thus, the remainder in this case is $- {\pi ^3} + 3{\pi ^2} - 3\pi + 1.$
$(v)$ When the polynomial is divided by $5 + 2x$, we can find the remainder by putting $5 + 2x = 0$ i.e. $x = - \dfrac{5}{2}$ in the polynomial:
$\Rightarrow {\text{Remainder}} = {( - \dfrac{5}{2})^3} + 3{( - \dfrac{5}{2})^2} + 3( - \dfrac{5}{2}) + 1, \\ \Rightarrow {\text{Remainder}} = - \dfrac{{125}}{8} + \dfrac{{75}}{4} - \dfrac{{15}}{2} + 1, \\ \Rightarrow {\text{Remainder}} = \dfrac{{ - 125 + 150 - 60 + 8}}{8}, \\ \Rightarrow {\text{Remainder}} = - \dfrac{{27}}{8}. \\$
Thus, the remainder in this case is $- \dfrac{{27}}{8}.$

Note: If on dividing the polynomial by $x + a$ the remainder comes out as $0,$ then $x + a$ is the factor of the polynomial, i.e. the polynomial is perfectly divisible by $x + a.$