# Questions on remainder theorem of polynomial

Given below is the collection of questions related to remainder theorem of polynomial.

All the questions are to the standard of grade 09.

Question 01
Consider the two functions f(x) and g(x);

f(x) = \mathtt{\ \ x^{3} -6x^{2} +2x-4}

g(x) = 1 – 2x

Find the remainder when f(x) is divided by g(x).

Solution
Let’s first find the root of g(x)

1-2x = 0

1= 2x

x = 1/2

Now put this value in function f(x) to get the value of remainder.

\mathtt{\Longrightarrow \ \ x^{3} -6x^{2} +2x-4}\\\ \\ \mathtt{\Longrightarrow \ \ \left(\frac{1}{2}\right)^{3} -6\left(\frac{1}{2}\right)^{2} +2\left(\frac{1}{2}\right) -4}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{8} -\frac{6}{4} +1-4}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{8} -\frac{6}{4} +-3}\\\ \\ \mathtt{\Longrightarrow \ \frac{1-12-24}{8}}\\\ \\ \mathtt{\Longrightarrow \ -\frac{35}{8}}

Hence, -35/8 is the solution of given expression.

Question 02
Consider the below function f(x) and g(x)

f(x) = \mathtt{x^{4} \ –\ 3x^{2} \ +\ 4}

g(x) = x – 2

Find the remainder when f(x) is divided by g(x).

Solution
Let’s first calculate the root of g(x).

x – 2 = 0

x = 2

Put this value of x in function f(x) to get the remainder.

\mathtt{\Longrightarrow \ x^{4} \ –\ 3x^{2} \ +\ 4}\\\ \\ \mathtt{\Longrightarrow \ ( 2)^{4} \ –\ 3( 2)^{2} \ +\ 4}\\\ \\ \mathtt{\Longrightarrow \ 16-12+4}\\\ \\ \mathtt{\Longrightarrow 8}

Hence, 8 is the remainder.

Question 03
Given below are two functions f(x) and g(x).

f(x) = \mathtt{9x^{3} \ –\ 3x^{2} \ +\ x-5}

g(x) = x – 2/3

Find the remainder when f(x) is divided by g(x).

Solution
First find the root of function g(x).

x – 2/3 = 0

x = 2/3

Put this value of x in function f(x) to get the remainder.

\mathtt{\Longrightarrow \ 9x^{3} \ –\ 3x^{2} \ +\ x-5}\\\ \\ \mathtt{\Longrightarrow \ 9\left(\frac{2}{3}\right)^{3} \ –\ 3\left(\frac{2}{3}\right)^{2} \ +\ \frac{2}{3} -5}\\\ \\ \mathtt{\Longrightarrow \ 9\left(\frac{8}{27}\right) -3\left(\frac{4}{9}\right) +\frac{2}{3} -5}\\\ \\ \mathtt{\Longrightarrow \frac{8}{3} -\frac{4}{3} +\frac{2}{3} -5}\\\ \\ \mathtt{\Longrightarrow \frac{6}{3} -5}\\\ \\ \mathtt{\Longrightarrow 2-5}\\\ \\ \mathtt{\Longrightarrow \ -3}

Hence, -3 is the remainder.

Question 04
Given below are three polynomial f(x), g(x), k(x). When function f(x) & g(x) are divided by k(x), they leave the same remainder. Find the value of variable “a”.

f(x) = \mathtt{2x^{3} \ +\ ax^{2} \ +\ 3x\ –\ 5}

g(x) = \mathtt{x^{3} \ +\ x^{2} \ –\ 4x\ +\ a}

k(x) = x – 2

Solution
First find the root of function k(x).

x – 2 = 0

x = 2

Let’s find remainder when f(x) is divided by k(x)

\mathtt{\Longrightarrow \ 2x^{3} \ +\ ax^{2} \ +\ 3x\ –\ 5}\\\ \\ \mathtt{\Longrightarrow \ 2( 2)^{3} +a( 2)^{2} +3( 2) -5}\\\ \\ \mathtt{\Longrightarrow \ 2( 8) +4a+6-5}\\\ \\ \mathtt{\Longrightarrow \ 17+4a}

Now find the remainder when g(x) divided by k(x).

\mathtt{\Longrightarrow \ x^{3} \ +\ x^{2} \ –\ 4x\ +\ a}\\\ \\ \mathtt{\Longrightarrow \ ( 2)^{3} \ +\ ( 2)^{2} \ –\ 4( 2) \ +\ a}\\\ \\ \mathtt{\Longrightarrow \ 8+4-8+a}\\\ \\ \mathtt{\Longrightarrow \ 4+a}

It’s given that both the remainders are same.

\mathtt{4+\ a\ =\ 17\ +\ 4a}\\\ \\ \mathtt{3a\ =\ -17+4}\\\ \\ \mathtt{3a=\ -13}\\\ \\ \mathtt{a=\ \frac{-13}{3}}

Hence, value of a is -13/3.

Question 05
Find the remainder when \mathtt{x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1} is divided by 5+2x.

Solution
Finding the root of expression 5 + 2x

5 + 2x = 0

x = -5/2

Putting this value of x in given expression to find the remainder.

\mathtt{\Longrightarrow \ x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1}\\\ \\ \mathtt{\Longrightarrow \ \left(\frac{-5}{2}\right)^{3} \ +\ 3\left(\frac{-5}{2}\right)^{2} \ +\ 3\left(\frac{-5}{2}\right) \ +\ 1}\\\ \\ \mathtt{\Longrightarrow \ \frac{-125}{8} +3\left(\frac{25}{4}\right) -\frac{15}{2} +1}\\\ \\ \mathtt{\Longrightarrow \ \frac{-125+150-60+8}{8}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-185+158}{8}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-27}{8}}

Hence, -27/8 is the remainder.

Next chapter : Problems on remainder theorem of polynomial (set 02)