Factor completely.
-$20 x^{2}+31 x+12$
A) $(20 x+3)(x+4)$
B) $(4 x-3)(5 x-4)$
C) $(4 x+3)(5 x+4)$
D) $(20 x+1)(x+12)$

Question

Quizplus · Accepted Answer

The correct factorization of $20x^2 + 31x + 12$ is obtained by finding two numbers that multiply to $20 \times 12 = 240$ and add to $31$. Those numbers are $16$ and $15$, which can be used to rewrite the middle term: $20x^2 + 16x + 15x + 12$. Factoring by grouping, we get $4x(5x + 4) + 3(5x + 4)$, which simplifies to $(4x + 3)(5x + 4)$.

Factor Completely 20x2+31x+1220 x^{2}+31 x+1220x2+31x+12 A) (20x+3)(x+4)(20 x+3)(x+4)(20x+3)(x+4) B) (4x−3)(5x−4)(4 x-3)(5 x-4)(4x−3)(5x−4) C) (4x+3)(5x+4)(4 x+3)(5 x+4)(4x+3)(5x+4)

Factor Completely $20 x^{2}+31 x+12$ A) $(20 x+3)(x+4)$ B) $(4 x-3)(5 x-4)$ C) $(4 x+3)(5 x+4)$