How do you factor completely: $2 x^{2} - 6 x - 56$ $2x^2 - 6x - 56$ ?

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How do you factor completely: $2 x^{2} - 6 x - 56$ $2x^2 - 6x - 56$ ?

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Answer 1 · 2015-07-18T16:42:54

Separate out the scalar factor $2$ $2$ then find pair of numbers $4$ $4$ and $7$ $7$ whose product is $28$ $28$ and difference $3$ $3$ , hence...

$2 x^{2} - 6 x - 56 = 2 (x - 7) (x + 4)$ $2x^2-6x-56 = 2(x-7)(x+4)$

Explanation:

$2 x^{2} - 6 x - 56 = 2 (x^{2} - 3 x - 28)$ $2x^2-6x-56 = 2(x^2-3x-28)$

To factor $x^{2} - 3 x - 28$ $x^2-3x-28$ , find a pair of numbers whose product is $28$ $28$ and whose difference is $3$ $3$ . The pair $7$ $7$ , $4$ $4$ works.

So $x^{2} - 3 x - 28 = (x - 7) (x + 4)$ $x^2-3x-28 = (x-7)(x+4)$

and

$2 x^{2} - 6 x - 56 = 2 (x - 7) (x + 4)$ $2x^2-6x-56 = 2(x-7)(x+4)$

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How do you factor completely: $2 x^{2} - 6 x - 56$ $2x^2 - 6x - 56$ ?

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How do you factor completely: $2 x^{2} - 6 x - 56$ $2x^2 - 6x - 56$ ?