How do you factor $x^{2} + 11 x = 180$ $x^2+ 11x = 180$ ?

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How do you factor $x^{2} + 11 x = 180$ $x^2+ 11x = 180$ ?

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Answer 1 · 2015-05-24T09:01:01

First subtract $180$ $180$ from both sides to get:

$x^{2} + 11 x - 180 = 0$ $x^2+11x-180 = 0$

Noticing that the coefficient of $x^{2}$ $x^2$ is $1$ $1$ and the signs of the other two coefficients, this may factorize as:

$(x + a) (x - b) = x^{2} + (a - b) x - a b$ $(x+a)(x-b) = x^2+(a-b)x-ab$ with $a > 0$ $a > 0$ , $b > 0$ $b > 0$ , $a - b = 11$ $a-b = 11$ and $a b = 180$ $ab = 180$ .

$180 = 20 \times 9$ $180 = 20 xx 9$ and $11 = 20 - 9$ $11 = 20 - 9$

So we can let $a = 20$ $a=20$ and $b = 9$ $b=9$ to get:

$x^{2} + 11 x - 180 = (x + 20) (x - 9)$ $x^2+11x-180 = (x+20)(x-9)$

Hence $x^{2} + 11 x = 180$ $x^2+11x=180$ has solutions $x = - 20$ $x=-20$ and $x = 9$ $x=9$

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How do you factor $x^{2} + 11 x = 180$ $x^2+ 11x = 180$ ?

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