Square of a Binomial

The square of a binomial is always a trinomial.  It will be helpful to memorize these patterns for writing squares of binomials as trinomials.

${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$

${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$

Examples:

Square each binomial.

$a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(x+4\right)}^{2}$

$\begin{array}{l}{\left(x+4\right)}^{2}={x}^{2}+2\left(x\cdot 4\right)+{4}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={x}^{2}+8x+16\end{array}$

$b\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(2y-3\right)}^{2}$

$\begin{array}{l}{\left(2y-3\right)}^{2}={\left(2y\right)}^{2}-2\left(2y\cdot 3\right)+{3}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(2y\right)}^{2}-2\left(6y\right)+{3}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4{y}^{2}-12y+9\end{array}$

$c\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(3p-2{q}^{2}\right)}^{2}$

$\begin{array}{l}{\left(3p-2{q}^{2}\right)}^{2}={\left(3p\right)}^{2}-2\left(3p\cdot 2{q}^{2}\right)+{\left(2{q}^{2}\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=9{p}^{2}-2\left(6p{q}^{2}\right)+4{q}^{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=9{p}^{2}-12p{q}^{2}+4{q}^{4}\end{array}$

If the coefficients of a trinomial $a{x}^{2}+bx+c$ satisfy the equation

$c={\left(\frac{b}{2\sqrt{a}}\right)}^{2}$ ,

then the trinomial is the perfect square of the binomial

$\sqrt{a}x+\frac{b}{2}$ .

Example 1:

Factor, if possible.

${x}^{2}-14x+49$

Here, $a=1,b=-14,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=49$ . We have:

${\left(\frac{b}{2\sqrt{a}}\right)}^{2}={\left(\frac{-14}{2\sqrt{1}}\right)}^{2}={\left(-7\right)}^{2}=49=c$

So, the trinomial is a perfect square:

${x}^{2}-14x+49={\left(x-7\right)}^{2}$

You can verify this using FOIL .

Example 2:

Factor, if possible.

$9{w}^{4}+12{w}^{2}+4$

Here, $a=9,b=12,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=4$ . (We can treat ${w}^{2}$ as $x$ , and not worry about the fourth power.)

${\left(\frac{b}{2\sqrt{a}}\right)}^{2}={\left(\frac{12}{2\sqrt{9}}\right)}^{2}={\left(\frac{12}{6}\right)}^{2}=4=c$

So, the trinomial is a perfect square:

$9{w}^{4}+12{w}^{2}+4={\left(3{w}^{2}+2\right)}^{2}$

This can also be verified using FOIL.