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When working with fractions, it is sometimes practical to put every fraction on the same denominator. Finding the common denominator is important when comparing fractions, ordering fractions, or performing mathematical operations such as adding and subtracting fractions. In each case, we refer to the common denominator. The following are a few methods for finding a common denominator in two or more fractions.
When the denominators of the fractions are algebraic expressions, the method for determining a common denominator is similar to the factor tree method above.
Factor and reduce each of the fractions.
Determine the common denominator.
Find the equivalent fractions.
Because of its similarity with the factor tree, there is an emphasis on factoring. Thus, it is essential to master the different methods of factoring a polynomial.
Find the common denominator of the following fractions:
||\frac{3x^2+6x}{x^2+5x+6} \ \ \text{and} \ \ \frac{2x-6}{6x^2+36x+54}||
1. Factor and reduce each of the fractions
||\begin{align} \small \frac{\color{blue}{3x^2+6x}}{\color{red}{x^2+5x+6}} &\Rightarrow \small \color{blue}{3x^2 + 6x} &&&& \small\color{red}{x^2+5x+6} \\
&= \small \color{blue}{3x(x+2)} && \small \text{factoring the GCF} && \small\color{red}{(x+3)(x+2)} && \small \text{product sum}\\
\small \frac{\color{blue}{3x^2+6x}}{\color{red}{x^2+5x+6}} &= \small \frac{\color{blue}{3x (x+2)}}{\color{red}{(x+3)(x+2)}} \\
&= \small \frac{\color{blue}{3x}}{\color{red}{(x+3)}} && \small \text{simplification}\\\\
\small \frac{\color{green}{2x-6}}{\color{orange}{6x^2+36x+54}} &\Rightarrow \small \color{green}{2x-6} &&&& \small\color{orange}{6x^2+36x+54} \\
&= \small \color{green}{2(x-3)} && \small \text{factoring the GCF} && \small\color{orange}{6(x^2+6x+9)} && \small \text{factoring the GCF}\\
&&&&& \small \color{orange}{6(x+3)(x+3)} && \small \text{perfect square}\\
\small \frac{\color{green}{2x-6}}{\color{orange}{6x^2+36x+54}} &= \small \frac{\color{green}{2(x-3)}}{\color{orange}{6(x+3)(x+3)}} \\
&= \small \frac{\color{green}{(x-3)}}{\color{orange}{3(x+3)(x+3)}} && \small \text{simplification} \end{align}||
2. Determine the common denominator
In this step, ensure each denominator’s factors are found in the common denominator. If part of the first denominator is identical ("pairs") to part of the second denominator, only one of the "pairs" is kept.
||\begin{align} \small\text{denominator} &= \small \color{red}{(x+3)} && \small\text{and} && \small \color{orange}{3(x+3)(x+3)} \\
\small \text{common denominator} &= \small \underbrace{\color{red}{(x+3)}}_{\small\text{pairs}} \ \color{orange}{3} \ \underbrace{\color{orange}{(x+3)}}_{\small\text{pairs}} \ \color{orange}{(x+3)} && \small \text{factoring common denominators}\\
&= \small\underbrace{\color{red}{(x+3)}}_{\small\text{1 of the pairs}}\ \small\color{orange}{3} \phantom{(x+3)} \color{orange}{(x+3)} && \small \text{eliminating one of the “pairs”} \\
&= \small 3 \ (x+3) \ (x+3) && \small \text{common denominator} \end{align}||
3. Find the equivalent fractions
Finally, multiply the numerators and denominators of the initial fractions by the missing elements of the common denominator |\small 3 \ (x+3) \ (x+3)| .
||\begin{align} \small \frac{\color{blue}{3x}}{\color{red}{(x+3)}} &\Rightarrow \small\frac{\color{blue}{3x}}{\underbrace{\color{red}{(x+3)}}_{\small\text{initial}}}\cdot \frac{3(x+3)}{\underbrace{3 \ (x+3)}_{\small\text{missing}}} && \small\underbrace{\phantom{(}3\phantom}_{\small\text{missing}}\small\underbrace{(x+3)}_{\small\text{common}}\ \ \small\underbrace{(x+3)}_{\small\text{missing}} \\
&= \small\frac{9x^2+27x}{3 (x+3)(x+3)} \\\\
\small \frac{\color{green}{(x-3)}}{\color{orange}{3(x+3)(x+3)}} &\Rightarrow \small \frac{\color{green}{(x-3)}}{\underbrace{\color{orange}{3 \ (x+3) \ (x+3)}}_{\small\text{initial}}} \cdot \underbrace{\phantom{\frac{(\small\text{rien})}{(\small\text{rien})}}}_{\small\text{missing}} && \small\underbrace{3 \ (x+3) \ (x+3)}_{\small\text{common}} \\
&=\small \frac{(x-3)}{3\ (x+3)\ (x+3)} \end{align}||
Since the second initial fraction has no missing elements, it remains unchanged.
Thus,
||\begin{align} \small \frac{3x^2+6x}{x^2+5x+6}&&& \text{and} && \small \frac{2x-6}{6x^2+36x+54} \\\\
\Rightarrow \small\frac{9x^2+27x}{3 (x+3)(x+3)} &&& \text{and} && \small \frac{(x-3)}{3(x+3)(x+3)} \end{align}||
The two fractions now have a common denominator, therefore, we can add or subtract them.