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


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


Division Algorithm for Polynomials

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


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


Division Algorithm for Polynomials

        \    First polynomial is a factor of second polynomial.


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.


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).


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?


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

Division Algorithm for Polynomials

    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.


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?


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)

Division Algorithm for Polynomials

            \displaystyle r(x)=-5x+8

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


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.


\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

Division Algorithm for Polynomials

        \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


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)


\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}

Division Algorithm for Polynomials

\    \displaystyle g(x)={{x}^{2}}-x+1


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}.


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,

Division Algorithm for Polynomials

        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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