Factor by grouping.
-$$8 \mathrm { x } ^ { 2 } + 12 \mathrm { x } + 10 \mathrm { x } + 15$$
A) $( 4 x + 5 ) ( 2 x + 3 )$
B) $( 8 x + 5 ) ( x + 3 )$
C) $( 4 x - 5 ) ( 2 x - 3 )$
D) $( 8 x - 5 ) ( x - 3 )$

Question

Quizplus · Accepted Answer

First, group the terms: $8x^2 + 12x$ and $10x + 15$. Factoring out the greatest common factor from each group gives $4x(2x + 3) + 5(2x + 3)$. Then, factor out the common factor $(2x + 3)$, resulting in $(4x + 5)(2x + 3)$.

Factor by Grouping 8x2+12x+10x+158 \mathrm { x } ^ { 2 } + 12 \mathrm { x } + 10 \mathrm { x } + 158x2+12x+10x+15

Factor by Grouping $8 \mathrm { x } ^ { 2 } + 12 \mathrm { x } + 10 \mathrm { x } + 15$