How can we find the quotient and remainder when:

$f(x)={x}^{5}-{x}^{4}-4{x}^{3}+2x+3$

is divided by

$g(x)=x-2?$

$f(x)={x}^{5}-{x}^{4}-4{x}^{3}+2x+3$

is divided by

$g(x)=x-2?$

taumulurtulkyoy
2022-10-23
Answered

How can we find the quotient and remainder when:

$f(x)={x}^{5}-{x}^{4}-4{x}^{3}+2x+3$

is divided by

$g(x)=x-2?$

$f(x)={x}^{5}-{x}^{4}-4{x}^{3}+2x+3$

is divided by

$g(x)=x-2?$

You can still ask an expert for help

garbhaighzf

Answered 2022-10-24
Author has **13** answers

A practical method is as follow:

$\begin{array}{r}\\ {x}^{5}-{x}^{4}-4{x}^{3}+2x+3& & & & & x-2\\ {x}^{4}-4{x}^{3}+2x+3& & & & & {x}^{4}\\ -2{x}^{3}+2x+3& & & & & {x}^{3}\\ -4{x}^{2}+2x+3& & & & & -2{x}^{2}\\ -6x+3& & & & & -4x\\ -9& & & & & -6\end{array}$

so the quotient is ${x}^{4}+{x}^{3}-2{x}^{2}-4x-6$ and the remainder is -9.

and to explain the procedure of calculus: we divide the leading term ${x}^{5}$ of the dividend by the leading term x of the divisor we find ${x}^{4}$ and then we calculate:

${x}^{5}-{x}^{4}-4{x}^{3}+2x+3-{x}^{4}(x-2)={x}^{4}-4{x}^{3}+2x+3=R(x)$

and repeat the same calculus using R(x) as your new dividend until we find the remainder R(x) with degree less than the degree of the divisor x-2.

$\begin{array}{r}\\ {x}^{5}-{x}^{4}-4{x}^{3}+2x+3& & & & & x-2\\ {x}^{4}-4{x}^{3}+2x+3& & & & & {x}^{4}\\ -2{x}^{3}+2x+3& & & & & {x}^{3}\\ -4{x}^{2}+2x+3& & & & & -2{x}^{2}\\ -6x+3& & & & & -4x\\ -9& & & & & -6\end{array}$

so the quotient is ${x}^{4}+{x}^{3}-2{x}^{2}-4x-6$ and the remainder is -9.

and to explain the procedure of calculus: we divide the leading term ${x}^{5}$ of the dividend by the leading term x of the divisor we find ${x}^{4}$ and then we calculate:

${x}^{5}-{x}^{4}-4{x}^{3}+2x+3-{x}^{4}(x-2)={x}^{4}-4{x}^{3}+2x+3=R(x)$

and repeat the same calculus using R(x) as your new dividend until we find the remainder R(x) with degree less than the degree of the divisor x-2.

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