Answer
$(2y-5)(8y-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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$16y^{2}-46y+15 =...$
Always start by searching for a GCF ... (there are none other than 1).
1. $\quad ac=+240\qquad $
2. $\quad$sum = $-46 \quad$... factors: $-40$ and $-6$
3. $\quad 16y^{2}-46y+15 = (16y^{2}-40y)+(-6y+15)$
4. $\quad$... $= 8y(2y-5)+(-3)(2y-5)=(2y-5)(8y-3)$
Check by FOIL
$F:\quad 16y^{2}$
$O:\quad -6y$
$I:\quad -40y$
$L:\quad +15$
$(2y-5)(8y-3) $ = $16y^{2}-46y+15$
