Factor $9y^2-30y+25$.
The quadratic is the square of $3y$,  the constant term is the square of $-5$, and the linear term equals $2(3y)(-5)$, so we have $9y^2 -30y + 25 = \boxed{(3y - 5)^2}$.
(3y-5)^2
Level 3
Algebra
Problem: Factor $9y^2-30y+25$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To factor the quadratic expression, recognize that it is in the standard form \( ax^2 + bx + c \). Identify the coefficients: \( a = 9 \), \( b = -30 \), and \( c = 25 \). 

Next, calculate the discriminant using the formula \( D = b^2 - 4ac \). Here, \( D = (-30)^2 - 4 \cdot 9 \cdot 25 \). Compute \( D = 900 - 900 = 0 \). A discriminant of zero indicates that the quadratic has a perfect square factorization.

Since the discriminant is zero, the quadratic can be expressed as a square of a binomial. To find the binomial, use the formula for the roots of the quadratic \( y = \frac{-b \pm \sqrt{D}}{2a} \). With \( D = 0 \), the roots simplify to \( y = \frac{30}{18} = \frac{5}{3} \).

The quadratic can be rewritten using the vertex form, which is \( a(y - r)^2 \) where \( r \) is the root. Substitute \( r = \frac{5}{3} \) and \( a = 9 \) into the vertex form. This gives \( 9(y - \frac{5}{3})^2 \).

To express this in standard form, expand \( (y - \frac{5}{3})^2 \) to get \( y^2 - \frac{10}{3}y + \frac{25}{9} \). Multiply through by 9 to eliminate the fraction: \( 9y^2 - 30y + 25 \).

Thus, the expression \( 9y^2 - 30y + 25 \) factors as \( 9(y - \frac{5}{3})^2 \). Recognizing that \( 9 \) is \( (3)^2 \), rewrite it as \( (3(y - \frac{5}{3}))^2 \).

This simplifies to \( (3y - 5)^2 \). Therefore, the final answer is \( (3y - 5)^2 \).<|END_OF_LATENT|> The final answer is $(3y-5)^2$.