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Applications of the Algebraic Identities

Using the Algebraic Identities in Reverse

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 $a^{2} + 2 a b + b^{2}$ . This means that even though ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ , it’s just as important to remember the reverse, that $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ .

Example 1

Factorize $x^{2} + 6 x + 9$ 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 $a^{2} + 2 a b + b^{2}$ . 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 $2 a b = 6 x$ :

\begin{array}{l} \sqrt{9} & = \pm 3, \\ 3 \cdot 2 x & = 6 x, \\ - 3 \cdot 2 x & = - 6 x \end{array}

\sqrt{9} = \pm 3, 3 \cdot 2 x = 6 x, - 3 \cdot 2 x = - 6 x

You can see that

b

has to be 3, because

b = - 3

does not match with

2 a b = 6 x

. That gives you

\begin{array}{l} x^{2} + 6 x + 9 & = x^{2} + 2 \cdot 3 x + 3 \cdot 3 \\ = {(x + 3)}^{2} \end{array}

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

Example 2

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

First, you need to factor out the common factors:

\begin{array}{l} 2 x^{3} - 50 x \\ = 2 \cdot x \cdot x \cdot x - 2 \cdot 5 \cdot 5 \cdot x \\ = 2 x (x^{2} - 25) . \end{array}

\begin{array}{l} 2 x^{3} - 50 x & = 2 \cdot x \cdot x \cdot x - 2 \cdot 5 \cdot 5 \cdot x \\ = 2 x (x^{2} - 25) . \end{array}

The expression

(x^{2} - 25)

is of the form

(a^{2} - b^{2})

with

a = x

and

b = 5

, which means you can use the third algebraic identity in reverse. That gives you

(x^{2} - 25) = (x + 5) (x - 5)

, which in turn means that

2 x (x^{2} - 25) = 2 x (x + 5) (x - 5)

Simplifying Rational Expressions

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

\frac{x^{2} + a x + a^{2}}{x^{2} + b x + 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 $\frac{3 x - 6}{x^{2} - 4 x + 4}$

This fraction has two terms in the numerator and three terms in the denominator. The two terms in the numerator are $3 x = 3 \cdot x$ and $- 6 = - 2 \cdot 3$ . The three terms in the denominator are $x^{2} = x \cdot x$ , $- 4 x = - 4 \cdot x$ 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}{l} Numerator & = 3 x - 6 \\ = 3 (x - 2) \\ Denominator & = x^{2} - 4 x + 4 \\ = {(x - 2)}^{2} \\ = (x - 2) (x - 2) \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}{l} \frac{3 x - 6}{x^{2} - 4 x + 4} & = \frac{3 (x - 2)}{{(x - 2)}^{2}} \\ = \frac{3 (x-2)}{(x-2) (x - 2)} \\ = \frac{3}{x - 2} \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 $\frac{x + 1}{\frac{x^{2}}{4} + \frac{x}{2} + \frac{1}{4}}$

There’s nothing to be done with the numerator, but if you multiply the denominator by 4, you get

\begin{array}{l} 4 \cdot (\frac{x^{2}}{4} + \frac{x}{2} + \frac{1}{4}) & = x^{2} + 2 x + 1 \\ = {(x + 1)}^{2} . \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

\frac{x + 1}{\frac{x^{2}}{4} + \frac{x}{2} + \frac{1}{4}} = \frac{4 (x + 1)}{{(x + 1)}^{2}} = \frac{4}{x + 1}

The Third Algebraic Identity

Activities 2