How do you factor $6y^2 + 27y - 15$ ?

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Answer 1 · 2016-09-19T06:17:02

$6y^2+27y-15 = 3(2y-1)(y+5)$

First note that all of the coefficients are divisible by $3$ , so we can separate that out as a factor:

$6y^2+27y-15 = 3(2y^2+9y-5)$

We can factor the quadratic $2y^2+9y-5$ using an AC method:

Find a pair of factors of $AC = 2*5 = 10$ which differ by $B=9$

The pair $10, 1$ works.

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

$2y^2+9y-5 = (2y^2+10y)-(y+5)$

$color(white)(2y^2+9y-5) = 2y(y+5)-1(y+5)$

$color(white)(2y^2+9y-5) = (2y-1)(y+5)$

Putting it together:

$6y^2+27y-15 = 3(2y-1)(y+5)$

How do you factor 6y^2 + 27y - 15 6y^2 + 27y - 15?