Factor.
-$$y ^ { 2 } + 0.8 y + 0.16$$
A) $( y + 0.8 ) ^ { 2 }$
B) $\left( y ^ { 2 } + 0.4 \right) ^ { 2 }$
C) $( \mathrm { y } + 0.4 ) ^ { 2 }$
D) $y ( y + 0.8 )$

Question

Quizplus · Accepted Answer

This is a perfect square trinomial. The square root of $y^2$ is $y$, and the square root of $0.16$ is $0.4$. The middle term, $0.8y$, is twice the product of $y$ and $0.4$, confirming that the expression is a perfect square: $(y + 0.4)^2$.

Factor - y2+0.8y+0.16y ^ { 2 } + 0.8 y + 0.16y2+0.8y+0.16

Factor
- $y ^ { 2 } + 0.8 y + 0.16$