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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:

Division Algorithm for Polynomials

        \    \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:

Division Algorithm for Polynomials

        \    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

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.

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)

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).

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

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

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}

Division Algorithm for Polynomials

\    \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,

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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