X^2-3x-18/x+3? Simplify

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Answer 1 · 2018-05-20T00:43:32

$x + 6$ $x+6$

$\frac{x^{2} - 3 x - 18}{x + 3}$ $(x^2-3x-18)/(x+3)$

we need to factor first, the denominator will not factor but the numerator factors to:

$\frac{(x + 3) (x - 6)}{x + 3}$ $((x + 3) (x - 6))/(x+3)$

now cancel the factor $x + 3$ $x+3$ and we have the solution:

$x + 6$ $x+6$

Answer 2 · 2018-05-20T00:45:58

$\frac{x^{2} - 3 x - 18}{x + 3}$ $(x^2-3x-18)/(x+3)$ = $x - 6$ $x-6$

Firstly you need to factorize $x^{2} - 3 x - 18$ $x^2-3x-18$
$x^{2} - 3 x - 18$ $x^2-3x-18$ = $(x + 3) (x - 6)$ $(x+3)(x-6)$

Then you divide by $x + 3$ $x+3$
$\frac{(x + 3) (x - 6)}{x + 3}$ $[(x+3)(x-6)]/(x+3)$ = $x - 6$ $x-6$