How do you simplify $\frac{2 c^{2} + 2 c - 12}{- 8 + 2 c + c^{2}}$ $(2c^2+2c-12)/(-8+2c+c^2)$ ?

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How do you simplify $\frac{2 c^{2} + 2 c - 12}{- 8 + 2 c + c^{2}}$ $(2c^2+2c-12)/(-8+2c+c^2)$ ?

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Answer 1 · 2016-10-04T13:38:56

$\frac{2 (c + 3)}{c + 4}$ $(2(c+3))/(c+4)$

Explanation:

$\frac{2 c^{2} + 2 c - 12}{- 8 + 2 c + c^{2}}$ $(2c^2+2c-12)/(-8+2c+c^2)$

= $\frac{2 (c^{2} + c - 6)}{c^{2} + 2 c - 8} \frac{\leftarrow common factor of 2}{\leftarrow re-arrange the terms}$ $(2(c^2+c-6))/(c^2 +2c-8)" " (larr"common factor of 2")/(larr"re-arrange the terms")$

$=(2(c+3)cancel((c-2)))/((c+4)cancel((c-2))) " "larr$ find factors of quadratics

$=(2(c+3))/(c+4)$

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How do you simplify $\frac{2 c^{2} + 2 c - 12}{- 8 + 2 c + c^{2}}$ $(2c^2+2c-12)/(-8+2c+c^2)$ ?

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Explanation:

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How do you simplify (2c^2+2c-12)/(-8+2c+c^2)2c2+2c−12−8+2c+c2(2c^2+2c-12)/(-8+2c+c^2)?

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How do you simplify $\frac{2 c^{2} + 2 c - 12}{- 8 + 2 c + c^{2}}$ $(2c^2+2c-12)/(-8+2c+c^2)$ ?