Quotient-Remainder Theorem

The quotient remainder theorem states that when a linear polynomial \[q\left( x \right)\] with a zero \[x\text{ }=\text{ }a\] divides a polynomial \[p\left( x \right)\] (whose degree is greater than or equal to 1), the remainder is given by \[r\text{ }=\text{ }p\text{ }\left( a \right)\]. Without actually doing the long division steps, the quotient remainder theorem allows us to calculate the remainder of the division of any polynomial by a linear polynomial.

Table of Contents

• An Introduction to Quotient Remainder Theorem

• History of Etienne Bezout

• Statement of the Quotient Remainder Theorem

• Proof of the Quotient Remainder Theorem

• Applications of the Quotient Remainder Theorem

• Solved Examples

History of Etienne Bezout

Ettiene Bezout

Name: Ettiene Bezout

Born: 31 March 1930

Died: 27 September 1983

Contribution: He is credited with finding the remainder theorem in Polynomials

Statement of the Quotient Remainder Theorem

Given any integer $A$, and a positive integer $B$, there exist unique integers $Q$  and $R$  such that

$A=BQ+R$  where $0\le R\le B$

Proof of the Quotient Remainder Theorem

Let us assume that $q(x)$ and $r(x)$ are the quotient and the remainder, respectively. When a polynomial $p(x)$  is divided by a linear polynomial $(x-a)$,

By the division algorithm, $Dividend=Divisor\times Quotient+\operatorname{Re}mainder$

Using this, $p(x)=(x-a)q(x)+r(x)$

Substitute $x=a$ and $r(x)=r$

$p(a)=(a-a)q(a)+r$

$p(a)=0\times q(a)+r$

$p(a)=r$

Hence proved.

Applications of the Theorem

• This theorem helps us to get the quotient and the remainder without actually performing the long division method.

• This theorem also has some real-life applications.

Limitations of the theorem:

• The quotient remainder theorem doesn’t work properly with the polynomials having repeated roots.

• The remainder theorem does not work when the divisor is not linear

• It does not provide us with the quotient.

Solved Examples

1. Find the value of k if $\mathbf{p}\left( \mathbf{x} \right)\text{ }=\text{ }\left( \mathbf{3x}-\text{ }\text{ }\mathbf{2} \right)\left( \mathbf{x}-\text{ }\text{ }\mathbf{k} \right)\text{ }-\text{ }\mathbf{8}$  is divided by $\left( \mathbf{x}-\text{ }\text{ }\mathbf{2} \right)$ leaving the remainder $4$.

Ans: Given,

$p(x)=(3x-2)(x-k)-8$

Also, it is given that the remainder is $4$ when $p(x)$ is divided by $(x-2)$.

So, $a\text{ }=\text{ }2$  and $r\text{ }=\text{ }4$

Using the Remainder theorem,

$p(a)=r$

$p(2)=4$

$[3(2)-2](2-k)-8=4$

$(6-2)(2-k)=4+8$

$4(2-k)=12$

$k=2-3=-1$

The value of \[k\] is \[-1\]. 

2. If $\left( \mathbf{x}\text{ }-\text{ }\mathbf{8} \right)$ is one of the factors of $\mathbf{m}{{\mathbf{x}}^{\mathbf{3}}}~\text{}\mathbf{24}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{192x}\text{ }\text{ }\mathbf{512}$, find the value of m.

Ans: Let the given polynomial be $p(x)=m{{x}^{3}}-24{{x}^{2}}+192x-512$

Given that $\left( x\text{ }-\text{ }8 \right)$ is one of the factors of p(x):

So, \[r=0\] and \[a=8\].

By the Remainder theorem,

$p(a)=r$

$p(8)=0$

$m{{(8)}^{3}}-24{{(8)}^{2}}+192(8)-512=0$

$512m-1536+1536-512=0$

$512m=512$

$m=512/512=1$

The value of \[m\] is \[1\].

3. Find the remainder using the Remainder theorem for the following expression.

$({{\mathbf{x}}^{\mathbf{4}}}~-\text{}\mathbf{5}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }-\mathbf{2x}\text{ }+\text{ }\mathbf{6})\text{ }\div \text{ }\left( \mathbf{x}\text{ }+\text{ }\mathbf{4} \right)$.

Ans: Here, $p\left( x \right)\text{ }=({{\mathbf{x}}^{\mathbf{4}}}~-\text{}\mathbf{5}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }-\mathbf{2x}\text{ }+\text{ }\mathbf{6})$

And $x\text{ }+\text{ }4\text{ }=\text{ }x\text{ }-\text{ }\left( -4 \right)$

Comparing with $x\text{ }-\text{ }a$, we have $a\text{ }=\text{ }-4$

By the Remainder theorem,

$r=p(a)$

$r=\text{p}(-4)$

$r={{(-4)}^{4}}-5{{(-4)}^{3}}+{{(-4)}^{2}}-2(-4)+6$

$r=256+5(64)+16+8+6$

$r=256+320+16+8+6$

$r=606$

Hence, the remainder is $606$.

Summary

A remainder Theorem is an approach to the Euclidean division of polynomials. According to this theorem, if we divide a polynomial$ P(x)$ by a factor $( x – a)$, that isn’t essentially an element of the polynomial; you will find a smaller polynomial along with a remainder.

Important Formulas:

• Dividend = (Divisor × Quotient) + Remainder.

• When $P(x)$ is divided by $(x-a)$, remainder = $P(a)$

List of Related Articles

• Remainder Theorem

• Remainder Theorem and Polynomials

• What is the Remainder Theorem?