How do you factor $5d^2+6d-8$ ?

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Answer 1 · 2017-01-29T01:00:19

The factor of $5d^2+6d-8$ is

$(5d-4)(d+2)$

To factors the trinomial
$5d^2+6d-8$

We take the factors of the first coefficient $5 = 5*1$
We take the factors of the constant term $8 = 1*8 or 2*4$

We look for an arrangement of the factors that will subtract to get $+6$

$5*(+2) = 10$
$1 * (-4) = -4$
#10-4 = +6

Make $5d$ and $+2$ the outer factors
Make $1d$ and $-4$ the inner factors

$(5d-4)(d+2)$

How do you factor 5d^2+6d-85d^2+6d-8?