How do you divide $(x^4+2x^3-2x^2+9x+3)/(x^2+1)$ ?

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Answer 1 · 2018-03-24T09:12:05

$x^2+2x-3+(7x+6)/(x^2+1)$

By long division

$(x^4+2x^3-2x^2+9x+3):(x^2+1)=x^2$
$-x^4-x^2$

$(0+2x^3-3x^2+9x+3):(x^2+1)=x^2+2x$
$-2x^3-2x$

$(0+0-3x^2+7x+3):(x^2+1)=x^2+2x-3$
$+3x^2+3$

$(0+0+0+7x+6):(x^2+1)=x^2+2x-3$

$x^2+2x-3+(7x+6)/(x^2+1)$

How do you divide (x^4+2x^3-2x^2+9x+3)/(x^2+1) (x^4+2x^3-2x^2+9x+3)/(x^2+1) ?