Quotient-Remainder Theorem

The quotient remainder theorem states that when a linear polynomial $$q\left(x\right)$$ with a zero $$x=a$$ divides a polynomial $$p\left(x\right)$$ (whose degree is greater than or equal to 1), the remainder is given by $$r=p\left(a\right)$$. 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, $Dividend=Divisor\times Quotient+\mathrm{Re}mainder$

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 $\mathbf{p}\left(\mathbf{x}\right)=(\mathbf{3}\mathbf{x}-\mathbf{2})(\mathbf{x}-\mathbf{k})-\mathbf{8}$  is divided by $(\mathbf{x}-\mathbf{2})$ 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, $a=2$  and $r=4$

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

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

Given that $(x-8)$ 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.

$({\mathbf{x}}^{\mathbf{4}}-\mathbf{5}{\mathbf{x}}^{\mathbf{3}}+{\mathbf{x}}^{\mathbf{2}}-\mathbf{2}\mathbf{x}+\mathbf{6})÷(\mathbf{x}+\mathbf{4})$.

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

And $x+4=x-(-4)$

Comparing with $x-a$, we have $a=-4$

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?