factor the polynomial?

64y³ + 80y² + 25y =
y( 64y² + 80y + 25) =
y( 8y + 5)²

64y^3 + 80y^2 + 25y
= y(64y^2 + 80y + 25)
= y(64y^2 + 40y + 40y + 25)
= y[(64y^2 + 40y) + (40y + 25)]
= y[8y(8y + 5)) + 5(8y + 5)]
= y(8y + 5)(8y + 5)
= y(8y + 5)^2
(answer c)

c

you don't really have to factor this during a test
use the process of elimination, which is easy here

first, factor the y out, and we're left with 64y² + 80y + 25

a. NOT! because 5 * (-5) = - 25 but we need + 25
b. NOT! because 64y * 25 will give us a huge number of y terms
c. YES! 8y * 8y gives us 64y² 40y + 40y gives us 80y and 5 * 5 gives us the 25 term
next

y(8y+5)^2
c is correct.

y(64y² + 80y + 25)
y(8y + 5)(8y + 5)
y(8y + 5)²

64y^3 + 80y^2 + 25y
y(64y^2 + 80y + 25)
y(8y^2 + 5)^2

Answer is C.