How do you long divide $(x^4+10x^3+8x^2-59x+40) div (x^2+3x-5)$ ?

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Answer 1 · 2017-02-28T10:15:52

The remainder is $=0$ and the quotient is $=(x^2+7x-8)$

Let's perform the long division

$color(white)(aaaa)$ $x^4+10x^3+8x^2-59x+40$ $color(white)(aaaa)$ $|$ $x^2+3x-5$

$color(white)(aaaa)$ $x^4+3x^3-5x^2$ $color(white)(aaaaaaaaaaaaaaa)$ $|$ $x^2+7x-8$

$color(white)(aaaaa)$ $0+7x^3+13x^2-59x$

$color(white)(aaaaaaa)$ $+7x^3+21x^2-35x$

$color(white)(aaaaaaaaa)$ $+0-8x^2-24x+40$

$color(white)(aaaaaaaaaaaaa)$ $-8x^2-24x+40$

$color(white)(aaaaaaaaaaaaaaaa)$ $-0-0+0$

The remainder is $=0$ and the quotient is $=(x^2+7x-8)$

How do you long divide (x^4+10x^3+8x^2-59x+40) div (x^2+3x-5) (x^4+10x^3+8x^2-59x+40) div (x^2+3x-5)?