How do you factor $2 x^{2} + 7 x - 4$ $2x^2 + 7x-4$ ?

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How do you factor $2 x^{2} + 7 x - 4$ $2x^2 + 7x-4$ ?

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Answer 1 · 2016-08-28T15:56:21

$2 x^{2} + 7 x - 4 = (2 x - 1) (x + 4)$ $2x^2+7x-4=(2x-1)(x+4)$

Explanation:

Use an AC method:

Find a pair of factors of $A C = 2 \cdot 4 = 8$ $AC=2*4=8$ which differ by $B = 7$ $B=7$

The pair $8, 1$ $8, 1$ works.

Use this pair to split the middle term and factor by grouping:

$2 x^{2} + 7 x - 4$ $2x^2+7x-4$

$= 2 x^{2} + 8 x - x - 4$ $=2x^2+8x-x-4$

$= (2 x^{2} + 8 x) - (x + 4)$ $=(2x^2+8x)-(x+4)$

$= 2 x (x + 4) - 1 (x + 4)$ $=2x(x+4)-1(x+4)$

$= (2 x - 1) (x + 4)$ $=(2x-1)(x+4)$

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How do you factor $2 x^{2} + 7 x - 4$ $2x^2 + 7x-4$ ?

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