Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2x-3}{8x^3-27}(2x^2+6x+9)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{4x^2+6x+9}(2x^2+6x+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2x^2+6x+9}{4x^2+6x+9}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x-3}{8x^3-27} $ to $ \dfrac{1}{4x^2+6x+9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x-3}$. $$ \begin{aligned} \frac{2x-3}{8x^3-27} & =\frac{ 1 \cdot \color{blue}{ \left( 2x-3 \right) }}{ \left( 4x^2+6x+9 \right) \cdot \color{blue}{ \left( 2x-3 \right) }} = \\[1ex] &= \frac{1}{4x^2+6x+9} \end{aligned} $$ |
| ② | Multiply $ \dfrac{1}{4x^2+6x+9} $ by $ 2x^2+6x+9 $ to get $ \dfrac{ 2x^2+6x+9 }{ 4x^2+6x+9 } $. Step 1: Write $ 2x^2+6x+9 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{4x^2+6x+9} \cdot 2x^2+6x+9 & \xlongequal{\text{Step 1}} \frac{1}{4x^2+6x+9} \cdot \frac{2x^2+6x+9}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( 2x^2+6x+9 \right) }{ \left( 4x^2+6x+9 \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^2+6x+9 }{ 4x^2+6x+9 } \end{aligned} $$ |