Solve (a-4)(a-4): Expanding a Repeated Binomial Expression

Because it comes from squaring a binomial! When you expand $(a-4)^2$, you get three terms that form a perfect pattern. The first and last terms are perfect squares, and the middle term is twice the product of the two original terms.

Yes! FOIL gives the same answer: First $a \cdot a = a^2$, Outer $a(-4) = -4a$, Inner $(-4)a = -4a$, Last $(-4)(-4) = 16$. Then combine: $a^2 - 4a - 4a + 16 = a^2 - 8a + 16$.

The middle term sign changes! $(a-4)^2 = a^2 - 8a + 16$ while $(a+4)^2 = a^2 + 8a + 16$. The first and last terms stay the same, but the middle term follows the original sign.

Think "First, Twice, Last": Square the first term, then twice the product of both terms, then square the last term. For $(a-4)^2$: First $a^2$, Twice $2(a)(-4) = -8a$, Last $(-4)^2 = 16$.

Because negative times negative equals positive! When you multiply $(-4) \times (-4) = +16$. This is why the last term in $(a-4)^2$ is always positive, even though we started with $-4$.