Factor completely.
-$10 y ^ { 2 } - 23 y + 12$
A) prime
B) $( 2 y - 3 ) ( 5 y - 4 )$
C) $( 2 y + 3 ) ( 5 y + 4 )$
D) $( 10 y + 3 ) ( y + 4 )$

Question

Quizplus · Accepted Answer

The correct factorization of $10y^2 - 23y + 12$ is found by looking for two numbers that multiply to $10 \times 12 = 120$ and add up to $-23$. These numbers are $-20$ and $-3$, which can be used to split the middle term and factor by grouping, leading to $(2y - 3)(5y - 4)$.

Factor Completely 10y2−23y+1210 y ^ { 2 } - 23 y + 1210y2−23y+12

Factor Completely $10 y ^ { 2 } - 23 y + 12$