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

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

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Answer 1 · 2017-03-01T00:01:22

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

Explanation:

First put in standard order of decreasing powers of $v$ $v$ :

$2 v^{2} - 7 v - 4$ $2v^2-7v-4$

Next, 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:

$2 v^{2} - 7 v - 4 = 2 v^{2} - 8 v + v - 4$ $2v^2-7v-4 = 2v^2-8v+v-4$

$2 v^{2} - 7 v - 4 = (2 v^{2} - 8 v) + (v - 4)$ $color(white)(2v^2-7v-4) = (2v^2-8v)+(v-4)$

$2 v^{2} - 7 v - 4 = 2 v (v - 4) + 1 (v - 4)$ $color(white)(2v^2-7v-4) = 2v(v-4)+1(v-4)$

$2 v^{2} - 7 v - 4 = (2 v + 1) (v - 4)$ $color(white)(2v^2-7v-4) = (2v+1)(v-4)$

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

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