Factor completely. If the polynomial cannot be factored, write ʺprime.ʺ
-$$49 y ^ { 4 } - 81$$
A) $\left( 7 y ^ { 2 } + 9 \right) ^ { 2 }$
B) prime
C) $\left( 7 y ^ { 2 } + 9 \right) \left( 7 y ^ { 2 } - 9 \right)$
D) $\left( 7 y ^ { 2 } - 9 \right) ^ { 2 }$

Question

Quizplus · Accepted Answer

Notice that this polynomial is in the form of a difference of squares: $49y^4-81=(7y^2)^2-9^2$. We can then factor using the difference of squares formula: $$(7y^2+9)(7y^2-9).$$ Simplifying, we get $$(7y^2+9)(7y^2-9)=(7y^2+9)(3y+3)(3y-3)=(7y^2+9)(3y+3)^2.$$ Since both factors cannot be further factored over the real numbers, the correct answer is $\boxed{\text{(C)}}$.

Factor Completely 49y4−8149 y ^ { 4 } - 8149y4−81 A) (7y2+9)2\left( 7 y ^ { 2 } + 9 \right) ^ { 2 }(7y2+9)2

Factor Completely $49 y ^ { 4 } - 81$ A) $\left( 7 y ^ { 2 } + 9 \right) ^ { 2 }$