The First and Second Algebraic Identity

You may have previously seen that ${\left(a×b\right)}^{2}={a}^{2}×{b}^{2}$, but what is ${\left(a+b\right)}^{2}$? If you write ${\left(a+b\right)}^{2}$ as $\left(a+b\right)\left(a+b\right)$, you can use what you learned about the distributive property of multiplying parentheses to find the answer:

$\begin{array}{llll}\hfill {\left(a+b\right)}^{2}& =\left(a+b\right)\left(a+b\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{2}+ab+ba+{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{2}+2ab+{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This also applies to ${\left(a-b\right)}^{2}$:

$\begin{array}{llll}\hfill {\left(a-b\right)}^{2}& =\left(a-b\right)\left(a-b\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{2}-ab-ba+{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{2}-2ab+{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

These results are used all the time in mathematics, and we call them the first and second algebraic identities. Learn these, and you’ll save a lot of time you would’ve spent calculating!

Rule

The First Algebraic Identity

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

The Second Algebraic Identity

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

Example 1

Expand ${\left(x-2\right)}^{2}+{\left(3+x\right)}^{2}$

If you apply the first and second algebraic identities to the squared parentheses, you get

$\begin{array}{llll}\hfill & \phantom{=}{\left(x-2\right)}^{2}+{\left(3+x\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-4x+4\right)+\phantom{\rule{-0.17em}{0ex}}\left(9+6x+{x}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{2}+2x+13\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {\left(x-2\right)}^{2}+{\left(3+x\right)}^{2}& =\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-4x+4\right)+\phantom{\rule{-0.17em}{0ex}}\left(9+6x+{x}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{2}+2x+13\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$