How do you factor completely $2 x^{2} + 7 x - 15$ $2x^2+7x-15$ ?

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How do you factor completely $2 x^{2} + 7 x - 15$ $2x^2+7x-15$ ?

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Answer 1 · 2017-02-10T00:11:52

$2 x^{2} + 7 x - 15 = (2 x - 3) (x + 5)$ $2x^2+7x-15 = (2x-3)(x+5)$

Explanation:

Given:

$2 x^{2} + 7 x - 15$ $2x^2+7x-15$

Use an AC method:

Find a pair of factors of $A C = 2 \cdot 15 = 30$ $AC=2*15=30$ which differ by $B = 7$ $B=7$ .

The pair $10, 3$ $10, 3$ works.

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

$2 x^{2} + 7 x - 15 = (2 x^{2} + 10 x) - (3 x + 15)$ $2x^2+7x-15 = (2x^2+10x)-(3x+15)$

$2 x^{2} + 7 x - 15 = 2 x (x + 5) - 3 (x + 5)$ $color(white)(2x^2+7x-15) = 2x(x+5)-3(x+5)$

$2 x^{2} + 7 x - 15 = (2 x - 3) (x + 5)$ $color(white)(2x^2+7x-15) = (2x-3)(x+5)$

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How do you factor completely $2 x^{2} + 7 x - 15$ $2x^2+7x-15$ ?

1 Answer

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How do you factor completely $2 x^{2} + 7 x - 15$ $2x^2+7x-15$ ?