Factor $36-9x^2$.
We have $36-9x^2 = 6^2 - (3x)^2 = (6-3x)(6+3x)$.  We can factor a 3 out of each of $6-3x$ and $6+3x$ to give $3\cdot(2-x)\cdot 3\cdot(2+x) = \boxed{9(2-x)(2+x)}$. (We could also have factored out a 9 at the beginning: $36-9x^2 = 9(4-x^2)=9(2-x)(2+x)$.)
9(2-x)(2+x)
Level 5
Algebra
Problem: Factor $36-9x^2$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To factor the expression, recognize that it is a difference of squares. The general form of a difference of squares is given by \( a^2 - b^2 = (a-b)(a+b) \). 

Identify \( a^2 \) and \( b^2 \) in the expression \( 36 - 9x^2 \). Here, \( 36 \) can be expressed as \( 6^2 \) and \( 9x^2 \) can be expressed as \( (3x)^2 \). 

Thus, we can rewrite the expression as \( 6^2 - (3x)^2 \). 

Now, apply the difference of squares formula: 

1. Set \( a = 6 \) and \( b = 3x \).
2. Substitute into the formula: \( (6 - 3x)(6 + 3x) \).

Next, simplify the factors. Notice that both factors can be further factored by taking out a common factor of 3 from each:

1. From \( 6 - 3x \), factor out 3 to get \( 3(2 - x) \).
2. From \( 6 + 3x \), factor out 3 to get \( 3(2 + x) \).

Now, combine the factored terms:

1. The expression becomes \( 3(2 - x) \cdot 3(2 + x) \).
2. This simplifies to \( 9(2 - x)(2 + x) \).

Thus, the final answer is \( 9(2 - x)(2 + x) \).<|END_OF_LATENT|> The final answer is $9(2-x)(2+x)$.