How do you factor $- x^{2} - 12 x - 18 = 0$ $-x^{2}-12x-18=0$ ?

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Answer 1 · 2018-04-22T14:29:37

The solution is $x = - 6 \pm 3 \sqrt{2}$ $x=-6±3sqrt2$

and hence the factorization is
$(x + 6 - 3 \sqrt{2}) (x + 6 + 3 \sqrt{2})) = 0$ $(x+6-3sqrt2)(x+6+3sqrt2))=0$

$- x^{2} - 12 x - 18 = 0$ $-x^2-12x-18=0$
$x^{2} + 12 x + 18 = 0$ $x^2+12x+18=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x= (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 12 \pm \sqrt{144 - 72}}{2 \times 1}$ $x= (-12+-sqrt(144-72))/(2 times1)$

$x = \frac{- 12 \pm \sqrt{72}}{2}$ $x=(-12+-sqrt72)/2$

$x = \frac{- 12 \pm 6 \sqrt{2}}{2}$ $x= (-12+-6sqrt2)/2$

$x = - 6 \pm 3 \sqrt{2}$ $x=-6±3sqrt2$

These are the solutions when $- x^{2} - 12 x - 18 = 0$ $-x^2-12x-18=0$

The factorization can be written in the form $(x - a) (x - b)$ $(x-a)(x-b)$ where $a$ $a$ and $b$ $b$ are the solutions, also known as zeros.

$(x - (- 6 + 3 \sqrt{2})) (x - (- 6 - 3 \sqrt{2})) = 0$ $(x-(-6+3sqrt2))(x-(-6-3sqrt2))=0$

$(x + 6 - 3 \sqrt{2}) (x + 6 + 3 \sqrt{2})) = 0$ $(x+6-3sqrt2)(x+6+3sqrt2))=0$

How do you factor -x^{2}-12x-18=0−x2−12x−18=0-x^{2}-12x-18=0?