How do you long divide $[(x^3)+(4x^2)-3x-12] div [(x^2)-3]$ ?

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Answer 1 · 2016-12-20T07:23:03

The remainder is $=0$ and the quotient is $(x+4)$

Let's do the long division

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

$color(white)(aaaa)$ $x^3$ $color(white)(aaaaaa)$ $-3x$ $color(white)(aaaaaaaaa)$ $∣$ $x+4$

$color(white)(aaaa)$ $0+4x^2$ $color(white)(aaaa)$ $0-12$

$color(white)(aaaaaa)$ $+4x^2$ $color(white)(aaaaaa)$ $-12$

$color(white)(aaaaaa)$ $+0$ $color(white)(aaaaaaaa)$ $-0$

The remainder is $=0$ and the quotient is $(x+4)$

How do you long divide [(x^3)+(4x^2)-3x-12] div [(x^2)-3][(x^3)+(4x^2)-3x-12] div [(x^2)-3]?