Answer
$2(2y-5)(y+3)$
Work Step by Step
Factoring by grouping:
1. Multiply the leading coefficient, a, and the constant, c.
2. Find the factors of ac whose sum is b.
3. Rewrite the middle term, bx, as a sum or difference using the factors from step 2.
4. Factor by grouping
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$4y^{2}+2y-30 =...$
Always start by searching for a GCF ... ($GCF=2$).
$=2(2y^{2}+y-15) =...$
Now, the parentheses:
1. $\quad ac=-30 \qquad $
2. $\quad$sum = $+1\quad$... factors: $-5$ and $+6$
3. $\quad (2y^{2}+y-15) = (2y^{2}+6y)+(-5y-15)$
4. $\quad$... $= 2y(y+3)+(-5)(y+3) = (2y-5)(y+3)$
$...$= $2(2y-5)(y+3)$
