How do you use long division to divide $(x^3+4x^2-3x-12)div(x-3)$ ?

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Answer 1 · 2017-04-12T12:54:54

The quotient is $=(x^2+7x+18)$ and the remainder is $=42$

Let's perform the long division

$color(white)(aaaa)$ $x^3+4x^2-3x-12$ $color(white)(aaaa)$ $|$ $x-3$

$color(white)(aaaa)$ $x^3-3x^2$ $color(white)(aaaaaaaaaaaaa)$ $|$ $x^2+7x+18$

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

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

$color(white)(aaaaaaaaaa)$ $0+18x-12$

$color(white)(aaaaaaaaaaaa)$ $+18x-54$

$color(white)(aaaaaaaaaaaaaa)$ $+0+42$

Therefore,

$(x^3+4x^2-3x-12)/(x-3)=x^2+7x+18+42/(x-3)$

The quotient is $=(x^2+7x+18)$ and the remainder is $=42$

How do you use long division to divide (x^3+4x^2-3x-12)div(x-3)(x^3+4x^2-3x-12)div(x-3)?