Find Equivalent Expressions for (x+3)²: Perfect Square Expansion

Because squaring a binomial means multiplying it by itself: $(x+3)(x+3)$. When you distribute, you get four terms that combine to $x^2 + 6x + 9$, not just the squares of each part!

Think "first, twice, last": square the first term, take twice the product of both terms, then square the last term. For $(x+3)^2$: first $x^2$, twice $2(x)(3) = 6x$, last $3^2 = 9$.

Absolutely! FOIL gives the same result. $(x+3)(x+3)$ becomes: First $x \cdot x = x^2$, Outer $x \cdot 3 = 3x$, Inner $3 \cdot x = 3x$, Last $3 \cdot 3 = 9$, then combine: $x^2 + 6x + 9$.

The same pattern applies! For $(x-3)^2$, you get $x^2 - 6x + 9$. The middle term takes the sign from the binomial: positive gives $+2ab$, negative gives $-2ab$.

Pick a simple value for x (like x = 1) and test both expressions. If $(1+3)^2 = 16$ and $1^2 + 6(1) + 9 = 16$, then your expansion is correct!