How do you long divide $(x^[4] + 3x^[3] + 28x + 15)/( x + 5)$ ?

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Answer 1 · 2017-01-24T07:03:40

The answer is $=(x^3-2x^2+10x-22)+125/(x+5)$

Let's do the long division

$color(white)(aaaa)$ $x^4+3x^3$ $color(white)(aaaaaaaa)$ $28x+15$ $color(white)(aaaa)$ $|$ $x+5$

$color(white)(aaaa)$ $x^4+5x^3$ $color(white)(aaaa)$ #color(white)(aaaaaaaaaaaaaaaa)|# $x^3-2x^2+10x-22$

$color(white)(aaaaa)$ $0-2x^3$

$color(white)(aaaaaaa)$ $-2x^3-10x^2$

$color(white)(aaaaaaaaa)$ $-0+10x^2+28x$

$color(white)(aaaaaaaaaaaaa)$ $+10x^2+50x$

$color(white)(aaaaaaaaaaaaaaaa)$ $+0-22x+15$

$color(white)(aaaaaaaaaaaaaaaaaaaa)$ $-22x-110$

$color(white)(aaaaaaaaaaaaaaaaaaaaaa)$ $+0+125$

Therefore,

$(x^4+3x^3+28x+15)/(x+5)=(x^3-2x^2+10x-22)+125/(x+5)$

How do you long divide (x^[4] + 3x^[3] + 28x + 15)/( x + 5)(x^[4] + 3x^[3] + 28x + 15)/( x + 5)?