It’s very useful to know what ${(a+b)}^{2}$ looks like when it has been expanded, but it’s often even more useful to know how to rewrite ${a}^{2}+2ab+{b}^{2}$. This means that even though ${(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$, it’s just as important to remember the reverse, that ${a}^{2}+2ab+{b}^{2}={(a+b)}^{2}$.

Example 1

**Factorize ${x}^{2}+6x+9$ by applying the first algebraic identity in reverse **

You want to end up with an expression of the form ${(a+b)}^{2}$, which means you need to find $a$ and $b$. The expression you have been given is of the form ${a}^{2}+2ab+{b}^{2}$. Because the first terms in the two expressions are ${a}^{2}$ and ${x}^{2}$, $a$ is equivalent to $x$.

So, you can find $b$ by setting ${b}^{2}=9$ and checking which values for $b$ yield $2ab=6x$:

$$\begin{array}{llll}\hfill \sqrt{9}& =\pm 3,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3\cdot 2x& =6x,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -3\cdot 2x& =-6x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

$$\sqrt{9}=\pm 3,\phantom{\rule{1em}{0ex}}3\cdot 2x=6x,\phantom{\rule{1em}{0ex}}-3\cdot 2x=-6x$$ |

Let’s look at an example where you also need to factor out common factors.

Example 2

**Factorize $2{x}^{3}-50x$ **

First, you need to factor out the common factors:

$$\begin{array}{llll}\hfill & \phantom{=}2{x}^{3}-50x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\cdot x\cdot x\cdot x-2\cdot 5\cdot 5\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

$$\begin{array}{llll}\hfill 2{x}^{3}-50x& =2\cdot x\cdot x\cdot x-2\cdot 5\cdot 5\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

The expression $\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right)$ is of the form $\phantom{\rule{-0.17em}{0ex}}\left({a}^{2}-{b}^{2}\right)$ with $a=x$ and $b=5$, which means you can use the third algebraic identity in reverse. That gives you $\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right)=(x+5)(x-5)$, which in turn means that $$2x\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}-25\right)=2x(x+5)(x-5)$$ |

If you apply everything you learned about factorization and algebraic identities, you can now simplify fractions that look like this:

$$\frac{{x}^{2}+ax+{a}^{2}}{{x}^{2}+bx+{b}^{2}}$$ |

When you simplify fractions, it’s extremely important to remember that common factors need to be present in all terms. If there’s a term in the fraction without a factor all the other terms have, you can’t cancel that factor! Here are some examples:

Example 3

**Simplify the fraction $\frac{3x-6}{{x}^{2}-4x+4}$ **

This fraction has two terms in the numerator and three terms in the denominator. The two terms in the numerator are $3x=3\cdot x$ and $-6=-2\cdot 3$. The three terms in the denominator are ${x}^{2}=x\cdot x$, $-4x=-4\cdot x$ and $4=2\cdot 2$. There are no factors that are common between all these terms, so you can’t cancel anything as-is.

But, if you factorize the numerator and denominator separately, you’ll see some common factors appearing: $$\begin{array}{llll}\hfill \text{Numerator}& =3x-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3(x-2)\phantom{\rule{2em}{0ex}}& \hfill & \\ \multicolumn{4}{c}{\text{}}\\ \phantom{\rule{2em}{0ex}}\\ \hfill \text{Denominator}& ={x}^{2}-4x+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={(x-2)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =(x-2)(x-2)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Now that you’ve factorized the numerator and denominator, you’re left with one term in each: $3(x-2)$ in the numerator and ${(x-2)}^{2}$ in the denominator. You can see that $(x-2)$ is a common factor between the two terms, which means it can be canceled: $$\begin{array}{llll}\hfill \frac{3x-6}{{x}^{2}-4x+4}& =\frac{3(x-2)}{{(x-2)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3\text{(x\u22122)}}{\text{(x\u22122)}(x-2)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3}{x-2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

**Note!** As you can see in this example, a term doesn’t have to be a single variable. It can also be an expression, such as $(x-2)$.

Example 4

**Simplify the fraction $\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}x+1\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{{x}^{2}}{4}+\frac{x}{2}+\frac{1}{4}\phantom{\rule{0.17em}{0ex}}}$ **

There’s nothing to be done with the numerator, but if you multiply the denominator by 4, you get $$\begin{array}{llll}\hfill 4\cdot \left(\frac{{x}^{2}}{4}+\frac{x}{2}+\frac{1}{4}\right)& ={x}^{2}+2x+1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={(x+1)}^{2}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

This new denominator is much easier to work with. Because you multiplied the denominator by 4, you need to do the same with the numerator. That means the numerator becomes $4\cdot (x+1)$. You can see that $(x+1)$ is a common factor, which gives you

$$\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}x+1\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{{x}^{2}}{4}+\frac{x}{2}+\frac{1}{4}\phantom{\rule{0.17em}{0ex}}}=\frac{4(x+1)}{{(x+1)}^{2}}=\frac{4}{x+1}$$ |

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