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

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Answer 1 · 2018-05-20T00:43:32

$x+6$

$(x^2-3x-18)/(x+3)$

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

$((x + 3) (x - 6))/(x+3)$

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

$x+6$

Answer 2 · 2018-05-20T00:45:58

$(x^2-3x-18)/(x+3)$ = $x-6$

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

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