How do you factor $x^{2} + 4 x - 12$ $x^2+4x-12$ ?

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How do you factor $x^{2} + 4 x - 12$ $x^2+4x-12$ ?

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Answer 1 · 2015-05-02T02:07:16

The answer is $(x - 2) (x + 6)$ $(x-2)(x+6)$

Determine what two numbers when added together make the coefficient of $x$ $x$ , in this case $4$ $4$ , and when multiplied together make the constant, in this case $- 12$ $-12$ . For this problem, the two numbers are $6$ $6$ and $- 2$ $-2$ .

$6 + (- 2) = 4$ $6+(-2)=4$ , and $6 \times (- 2) = 12$ $6xx(-2)=12$ .

The factorization of $x^{2} + 4 x - 12 = (x - 2) (x + 6)$ $x^2+4x-12=(x-2)(x+6)$ .

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How do you factor $x^{2} + 4 x - 12$ $x^2+4x-12$ ?

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