Question
Answer
$$\dfrac{18x^3-6x^2}{54x^4-6x^2}=\dfrac{1}{3x+1}$$
Explanation
| $$ \begin{aligned}\frac{18x^3-6x^2}{54x^4-6x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{3x+1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{18x^3-6x^2}{54x^4-6x^2} $ to $ \dfrac{1}{3x+1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{18x^3-6x^2}$. $$ \begin{aligned} \frac{18x^3-6x^2}{54x^4-6x^2} & =\frac{ 1 \cdot \color{blue}{ \left( 18x^3-6x^2 \right) }}{ \left( 3x+1 \right) \cdot \color{blue}{ \left( 18x^3-6x^2 \right) }} = \\[1ex] &= \frac{1}{3x+1} \end{aligned} $$ |
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Simplify Rational Expressions