Factor completely. $2 y ^ { 4 } - 5 y ^ { 2 } - 33$
A) $\left( 2 y ^ { 2 } - 11 \right) \left( y ^ { 2 } + 3 \right)$
B) $\left( 2 y ^ { 2 } + 11 \right) \left( y ^ { 2 } - 3 \right)$
C) $( 2 y + 11 ) \left( y ^ { 2 } - 3 \right)$
D) $\left( 2 y ^ { 2 } + 1 \right) \left( y ^ { 2 } - 33 \right)$
E) $\left( 2 y ^ { 2 } + 3 \right) \left( y ^ { 2 } - 11 \right)$

Question

Quizplus · Accepted Answer

We can factor this polynomial using grouping:
\begin{align*}
2y^4-5y^2-33 &= 2y^4-11y^2+6y^2-33 \
&= y^2(2y^2-11)+3(2y^2-11) \
&= (2y^2-11)(y^2+3)
\end{align*}
Therefore, the answer is choice A.

Factor Completely 2y4−5y2−332 y ^ { 4 } - 5 y ^ { 2 } - 332y4−5y2−33

Factor Completely $2 y ^ { 4 } - 5 y ^ { 2 } - 33$