Factor.
-$$y ^ { 2 } + 1.6 y + 0.64$$
A) $\left( y ^ { 2 } + 0.8 \right) ^ { 2 }$
B) $( \mathrm { y } + 1.6 ) ^ { 2 }$
C) $( y + 0.8 ) ^ { 2 }$
D) $y ( y + 1$.6

Question

Quizplus · Accepted Answer

We can factor the given expression using the perfect square trinomial formula, which states that if we have an expression of the form $ax^2+bx+c$ that can be written as $(mx+n)^2$, then $a=m^2$, $b=2mn$, and $c=n^2$. Applying this formula to $y^2+1.6y+0.64$, we see that it can be written as $(y+0.8)^2$, so the answer is $\boxed{	extbf{(C)}}$.

Factor - y2+1.6y+0.64y ^ { 2 } + 1.6 y + 0.64y2+1.6y+0.64

Factor
- $y ^ { 2 } + 1.6 y + 0.64$