Factor by any method.
-$$20 x ^ { 4 } + 27 x ^ { 2 } + 9$$
A) $\left( 4 x ^ { 2 } + 1 \right) \left( 5 x ^ { 2 } + 9 \right)$
B) $\left( 20 x ^ { 2 } + 3 \right) \left( x ^ { 2 } + 3 \right)$
C) $\left( 4 x ^ { 2 } + 3 \right) \left( 5 x ^ { 2 } + 3 \right)$
D) $\left( 5 x ^ { 2 } - 3 \right) \left( 4 x ^ { 2 } - 3 \right)$

Question

Quizplus · Accepted Answer

The polynomial $20x^4 + 27x^2 + 9$ can be factored as $(4x^2 + 3)(5x^2 + 3)$ by looking for factors of $20x^4$ and $9$ that when multiplied give a middle term of $27x^2$. The correct factors that satisfy these conditions are $4x^2$ and $5x^2$ for the $x^2$ terms, and $3$ for the constant term, making option C the correct factorization.

Factor by Any Method 20x4+27x2+920 x ^ { 4 } + 27 x ^ { 2 } + 920x4+27x2+9

Factor by Any Method $20 x ^ { 4 } + 27 x ^ { 2 } + 9$