How do you divide $\frac{x^{2} + 12 x - 18}{x - 3}$ $(x^2+12x-18)/(x-3)$ ?

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Answer 1 · 2018-03-08T00:59:39

$x + 9$ $x + 9$

$x - 3$ $x-3$ divided into $x^{2} + 12 x - 18$ $x^2+12x-18$

First part of answer is $x$ $x$

$x - 3$ $x-3$ times $x$ $x$ is $x^{2} - 3 x$ $x^2-3x$

Subtract from $x^{2} + 12 x - 18$ $x^2+12x-18$ to get $9 x - 18$ $9x-18$

Second part will then be $9$ $9$

$x - 3$ $x-3$ times $9$ $9$ is $9 x - 18$ $9x-18$

So answer is $x + 9$ $x+9$

How do you divide (x^2+12x-18)/(x-3)x2+12x−18x−3(x^2+12x-18)/(x-3)?