Square of a Binomial

The two patterns

For a sum: $(a + b)^{2} = a^{2} + 2ab + b^{2}$. For a difference: $(a - b)^{2} = a^{2} - 2ab + b^{2}$. In both cases you get the first term squared, plus or minus twice the product of the two terms, plus the last term squared.

Note the last term is positive in both patterns — squaring $-b$ gives $+b^{2}$. Only the middle term picks up the minus sign.

Where the middle term comes from

FOIL $(a + b)(a + b)$: First gives $a^{2}$, Outer gives $ab$, Inner gives $ba$, Last gives $b^{2}$. The Outer and Inner products are equal, and together they make $2ab$. That is why the middle term is always twice the product of the two terms — the pattern is just FOIL with the middle pair pre-combined.

So for $(x + 5)^{2}$: first term squared is $x^{2}$, twice the product is $2 \cdot x \cdot 5 = 10x$, last term squared is $25$. Answer: $x^{2} + 10x + 25$.

When the first term has a coefficient

The pattern still works when $a$ is more than a bare variable — you just have to square all of it. For $(2x + 6)^{2}$, take $a = 2x$ and $b = 6$: $a^{2} = 4x^{2}$, $2ab = 2 \cdot 2x \cdot 6 = 24x$, and $b^{2} = 36$. The answer is $4x^{2} + 24x + 36$.

Squaring $2x$ means squaring the $2$ as well: $(2x)^{2} = 4x^{2}$, not $2x^{2}$. That coefficient is the second-most-common slip after the missing middle term.

Worked examples

Try one yourself

Common questions