How do you factor completely $12 z^{2} + 7 z - 12$ $12z^2+7z-12$ ?

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How do you factor completely $12 z^{2} + 7 z - 12$ $12z^2+7z-12$ ?

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Answer 1 · 2018-05-19T11:54:25

See below

Explanation:

Two alternatives: use quadratic formula or complete square (a little bit complicate in this case). The use of Riffini's method is also available but in this case is not quite evident

Quadratic formula:

$12 z^{2} + 7 z - 12 = 0$ $12z^2+7z-12=0$

$z = \frac{- 7 \pm \sqrt{49 + 576}}{24} = \frac{- 7 \pm 25}{24}$ $z=(-7+-sqrt(49+576))/24=(-7+-25)/24$ from here two solutions

$z = \frac{3}{4}$ $z=3/4$ and $z = - \frac{4}{3}$ $z=-4/3$

Then factoring is $(z - \frac{3}{4}) (z + \frac{4}{3}) = 12 z^{2} + 7 z - 12$ $(z-3/4)(z+4/3)=12z^2+7z-12$

Answer 2 · 2018-05-19T13:11:14

$(3 z + 4) \cdot (4 z - 3)$ $(3z+4)*(4z-3)$

Explanation:

$12 z^{2} + 7 z - 12$ $12z^2+7z-12$

= $12 z^{2} + 16 z - 9 z - 12$ $12z^2+16z-9z-12$

= $4 z \cdot (3 z + 4) - 3 \cdot (3 z + 4)$ $4z*(3z+4)-3*(3z+4)$

= $(3 z + 4) \cdot (4 z - 3)$ $(3z+4)*(4z-3)$

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How do you factor completely $12 z^{2} + 7 z - 12$ $12z^2+7z-12$ ?

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