Factor completely.
-$12 x^{4}+17 x^{2}+6$
A) $\left(3 x^{2}+2\right)\left(4 x^{2}+3\right)$
B) $\left(3 x^{2}+1\right)\left(4 x^{2}+6\right)$
C) $\left(12 x^{2}+2\right)\left(x^{2}+3\right)$
D) $\left(4 x^{2}-3\right)\left(3 x^{2}-2\right)$

Question

Quizplus · Accepted Answer

The correct factorization of $12x^4 + 17x^2 + 6$ is obtained by finding two binomials whose product gives the original trinomial. The coefficients of $x^2$ in the binomials must multiply to give $12x^4$ (the first term) and their constant terms must multiply to give $6$ (the last term), while also summing in a way to give the middle term $17x^2$. The combination that works is $(3x^2 + 2)$ and $(4x^2 + 3)$, which when multiplied out, give the original polynomial.

Factor Completely 12x4+17x2+612 x^{4}+17 x^{2}+612x4+17x2+6 A) (3x2+2)(4x2+3)\left(3 x^{2}+2\right)\left(4 x^{2}+3\right)(3x2+2)(4x2+3) B) (3x2+1)(4x2+6)\left(3 x^{2}+1\right)\left(4 x^{2}+6\right)(3x2+1)(4x2+6)

Factor Completely $12 x^{4}+17 x^{2}+6$ A) $\left(3 x^{2}+2\right)\left(4 x^{2}+3\right)$ B) $\left(3 x^{2}+1\right)\left(4 x^{2}+6\right)$