Factor completely.
-$15 y^{2}+22 y+8$
A) $(15 y+2)(y+4)$
B) Prime
C) $(3 y+2)(5 y+4)$
D) $(3 y-2)(5 y-4)$

Name: Factor completely. -$15 y^{2}+22 y+8$ A) $(15 y+2)(y+4)$ B) Prime C) $(3 y+2)(5 y+4)$ D) $(3 y-2)(5 y-4)$
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Question

Quizplus · Accepted Answer

The quadratic $15y^2 + 22y + 8$ can be factored as $(3y+2)(5y+4)$ by finding two numbers that multiply to $15 \times 8 = 120$ and add to $22$, which are $10$ and $12$. Splitting the middle term and factoring by grouping gives the correct factorization.

Factor Completely 15y2+22y+815 y^{2}+22 y+815y2+22y+8 A) (15y+2)(y+4)(15 y+2)(y+4)(15y+2)(y+4) B) Prime C) (3y+2)(5y+4)(3 y+2)(5 y+4)(3y+2)(5y+4)

Factor Completely $15 y^{2}+22 y+8$ A) $(15 y+2)(y+4)$ B) Prime
C) $(3 y+2)(5 y+4)$