Use an algebraic identity to expand 4(x + 3)(x - 3).

Step-by-step solution

1. 1

   Spot the conjugate pair before multiplying anything. The two brackets differ only in the sign of the constant: $(x+3)$ and $(x-3)$. That is the signature of

   $(a+b)(a-b)=a^2-b^2$

   Using it beats FOIL because the two cross terms $+3x$ and $-3x$ are guaranteed to cancel, so there is nothing to track.

2. 2

   Apply the identity with $a=x$ and $b=3$.

   $(x+3)(x-3)=x^2-3^2=x^2-9$

   The constant is $3^2=9$, not $3$: the identity squares $b$.

3. 3

   Keep the leading factor of 4 outside for now. The 4 was never part of the identity, so it simply multiplies the result:

   $4(x+3)(x-3)=4(x^2-9)$

   This half-factored form $4(x^2-9)$ is often the form a textbook wants, since it shows the structure.

4. 4

   Distribute the 4 to finish the expansion.

   $4(x^2-9)=4x^2-36$

   Both terms get multiplied — writing $4x^2-9$ would be the classic slip.

5. 5

   Check at a convenient value. At $x=5$: the original is $4\cdot 8\cdot 2=64$, and $4(25)-36=100-36=64$. They match, so $4(x+3)(x-3)=4(x^2-9)=4x^2-36$. Read right to left, this also says $4x^2-36$ factors as $4(x+3)(x-3)$.