# CBSE 10th Mathematics | Division Algorithm for Polynomials

## Division Algorithm for Polynomials

Let $\displaystyle p(x)$ and $\displaystyle g(x)$ be polynomials of degree n and m respectively such that m £ n. Then there exist unique polynomials $\displaystyle q(x)$ and $\displaystyle r(x)$ where $\displaystyle r(x)$ is either zero polynomial or degree of $\displaystyle r(x)<$ degree of $\displaystyle g(x)$ such that $\displaystyle p(x)=q(x)\cdot g(x)+r(x)$.

$\displaystyle p(x)$ is dividend, $\displaystyle g(x)$ is divisor.

$\displaystyle q(x)$ is quotient, $\displaystyle r(x)$ is remainder.

## Solved Examples based on Division Algorithm for Polynomials

### Example:

Apply the division algorithm to find the quotient and remainder on dividing $\displaystyle p(x)$ by $\displaystyle g(x)$ as given below.

$\displaystyle p(x)={{x}^{3}}-3{{x}^{2}}+5x-3$, $\displaystyle g(x)={{x}^{2}}-2$

### Solution:

\    $\displaystyle Q(x)=x-3$     Þ    $\displaystyle r(x)=7x-9$

### Example:

Check whether the first polynomial is a factor of 2nd polynomial by applying division algorithm.

$\displaystyle {{t}^{2}}-3$, $\displaystyle 2{{t}^{4}}+3{{t}^{3}}-2{{t}^{2}}-9t-12$

### Solution:

\    First polynomial is a factor of second polynomial.

### Example:

If $\displaystyle (x-a)$ is the factor of the polynomial $\displaystyle {{x}^{3}}-m{{x}^{2}}-2nax+n{{a}^{2}},$ prove that $\displaystyle a=m+n$ and $\displaystyle a\ne 0.$

### Solution:

Let $\displaystyle p(x)={{x}^{3}}-m{{x}^{2}}-2nax+n{{a}^{2}}$

Since $\displaystyle (x-a)$ is a factor of $\displaystyle p(x)$

Þ    $\displaystyle p(a)=0$            Þ    $\displaystyle {{a}^{3}}-m{{a}^{2}}-2na\,\,.\,\,a+n{{a}^{2}}=0$

Þ    $\displaystyle {{a}^{3}}-{{a}^{2}}m-{{a}^{2}}n=0$        Þ    $\displaystyle {{a}^{2}}[a-(m+n)]=0$

Þ    $\displaystyle a-(m+n)=0$

as    $\displaystyle a\ne 0$

\    $\displaystyle a=(m+n).$

### Example:

What must be subtracted from $\displaystyle p(x)=6{{x}^{4}}+7{{x}^{3}}+26{{x}^{2}}-25x+25$ so that the resulting polynomial is exactly divisible by $\displaystyle g(x)=3{{x}^{2}}-x+4?$

### Solution:

By division algorithm, we have

Dividend = Divisor ´ Quotient + Remainder

Þ Dividend – Remainder = Divisor ´ Quotient

Þ $\displaystyle p(x)-r(x)=g(x)\times q(x)$

On dividing $\displaystyle p(x)$ by $\displaystyle g(x),$ we get

Thus if we subtract $\displaystyle -30x-3$ from $\displaystyle 6{{x}^{4}}+7{{x}^{3}}+26{{x}^{2}}-25x+25,$ it will be divisible by $\displaystyle 3{{x}^{2}}-x+4.$

### Example:

What must be added to $\displaystyle p(x)=4{{x}^{4}}-5{{x}^{3}}-39{{x}^{2}}-46x-2$ so that the resulting polynomial is divisible by $\displaystyle g(x)=4{{x}^{2}}+7x+2?$

### Solution:

By division algorithm, we have

$\displaystyle p(x)=g(x)\times q(x)+r(x)$

Þ $\displaystyle p(x)-r(x)=g(x)\times q(x)$

Þ $\displaystyle p(x)+[-r(x)]=g(x)\times q(x)$

Clearly RHS is divisible by $\displaystyle g(x)$

\ LHS is also divisible by $\displaystyle g(x)$

Thus, if $\displaystyle -r(x)$ is added to $\displaystyle p(x),$ then the resulting polynomial becomes divisible by $\displaystyle g(x)$

$\displaystyle r(x)=-5x+8$

Hence, we should add $\displaystyle -r(x)=5x-8,$ so that the resulting polynomial is divisible by g(x).

### Example:

If two zeros of the polynomial $\displaystyle f(x)={{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35$ are $\displaystyle 2\pm \sqrt{3}$ find other zeros.

### Solution:

$\displaystyle 2+\sqrt{3}$ and $\displaystyle 2-\sqrt{3}$ are zeros of $\displaystyle f(x)$

\    $\displaystyle \left\{ x-(2+\sqrt{3}) \right\}\,\left\{ x-\left( 2-\sqrt{3} \right)\, \right\}=\left( x-2-\sqrt{3} \right)\,\left( x-2+\sqrt{3} \right)$

= $\displaystyle {{(x-2)}^{2}}-{{(\sqrt{3})}^{2}}$

= $\displaystyle {{x}^{2}}-4x+1$ is a factor of $\displaystyle f(x)$.

Let us divide $\displaystyle f(x)$ by $\displaystyle {{x}^{2}}-4x+1$

$\displaystyle f(x)=({{x}^{2}}-4x+1)({{x}^{2}}-2x-35)$

Hence, other two zeros of $\displaystyle f(x)$ are zeros of polynomial $\displaystyle {{x}^{2}}-2x-35$

= $\displaystyle {{x}^{2}}-2x-35$

= $\displaystyle {{x}^{2}}-7x+5x-35$

= $\displaystyle x(x-7)+5(x-7)$

= $\displaystyle (x+5)(x-7)$

Other two zeros are –5, 7

### Example:

On dividing a polynomial $\displaystyle f(x)={{x}^{3}}-3{{x}^{2}}+x+2$ by a polynomial $\displaystyle g(x)$ the quotient $\displaystyle q(x)$ and remainder $\displaystyle r(x)$ are $\displaystyle x-2$ and $\displaystyle -2x+4$ respectively. Find $\displaystyle g(x)$

### Solution:

$\displaystyle f(x)=g(x)\cdot q(x)+r(x)$

$\displaystyle \frac{f(x)-r(x)}{q(x)}=g(x)$

$\displaystyle g(x)=\frac{{{x}^{3}}-3{{x}^{2}}+x+2-(-2x+4)}{x-2}$

=    $\displaystyle \frac{{{x}^{3}}-3{{x}^{2}}+x+2+2x-4}{x-2}=\frac{{{x}^{3}}-3{{x}^{2}}+3x-2}{x-2}$

\    $\displaystyle g(x)={{x}^{2}}-x+1$

### Example:

Find all the zeros of $\displaystyle 2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ if you know that two of its zeros are $\displaystyle \sqrt{2}$ and $\displaystyle -\sqrt{2}.$

### Solution:

Given two zeros : $\displaystyle \sqrt{2}$ and $\displaystyle \sqrt{-2}$

Þ    Quadratic polynomial = $\displaystyle (x+\sqrt{2})\,\,(x-\sqrt{2})$

$\displaystyle ={{x}^{2}}-{{(\sqrt{2})}^{2}}={{x}^{2}}-2$

Þ    $\displaystyle {{x}^{2}}-2$ is a factor of $\displaystyle 2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$

Applying the division algorithm theorem to given polynomial

$\displaystyle 2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ and $\displaystyle {{x}^{2}}-2,$

Clearly, we have

$\displaystyle 2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2=(2{{x}^{2}}-3x+1)\,\,({{x}^{2}}-2)$                         $\displaystyle =[2{{x}^{2}}-2x-x+1]\,({{x}^{2}}-2)=[2x(x-1)-1(x-1)]\,({{x}^{2}}-2)$                 $\displaystyle =(x-1)\,(2x-1)\,(x-\sqrt{2})\,(x+\sqrt{2})$

Therefore, all the zeros are : $\displaystyle 1,\frac{1}{2},\sqrt{2}$ and $\displaystyle -\sqrt{2}$

Þ    other two zeros are : $\displaystyle 1,\frac{1}{2}.$

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