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Answer

$$12x^2-12\cdot\frac{x}{8}x^2+8x=\frac{-12x^3+96x^2+64x}{8}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}12x^2-12 \cdot \frac{x}{8}x^2+8x& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(12-12 \cdot \frac{x}{8})x^2+8x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1-\frac{x}{8})\cdot12x^2+8x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-x+8}{8}\cdot12x^2+8x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-12x+96}{8}x^2+8x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-12x^3+96x^2}{8}+8x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-12x^3+96x^2+64x}{8}\end{aligned} $$
①
Use the distributive property.
②
Use the distributive property.
③

Subtract $ \dfrac{x}{8} $ from $ 1 $ to get $ \dfrac{ \color{purple}{ -x+8 } }{ 8 }$.

Step 1: Write $ 1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 8 }$.

$$ \begin{aligned} 1- \frac{x}{8} & \xlongequal{\text{Step 1}} \frac{1}{\color{red}{1}} - \frac{x}{8} = \frac{ 1 \cdot \color{blue}{ 8 }}{ 1 \cdot \color{blue}{ 8 }} - \frac{ x }{ 8 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 8 } }{ 8 } - \frac{ \color{purple}{ x } }{ 8 }=\frac{ \color{purple}{ -x+8 } }{ 8 } \end{aligned} $$
④

Multiply $ \dfrac{-x+8}{8} $ by $ 12 $ to get $ \dfrac{ -12x+96 }{ 8 } $.

Step 1: Write $ 12 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{-x+8}{8} \cdot 12 & \xlongequal{\text{Step 1}} \frac{-x+8}{8} \cdot \frac{12}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( -x+8 \right) \cdot 12 }{ 8 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -12x+96 }{ 8 } \end{aligned} $$
⑤

Multiply $ \dfrac{-12x+96}{8} $ by $ x^2 $ to get $ \dfrac{ -12x^3+96x^2 }{ 8 } $.

Step 1: Write $ x^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{-12x+96}{8} \cdot x^2 & \xlongequal{\text{Step 1}} \frac{-12x+96}{8} \cdot \frac{x^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( -12x+96 \right) \cdot x^2 }{ 8 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -12x^3+96x^2 }{ 8 } \end{aligned} $$
⑥

Add $ \dfrac{-12x^3+96x^2}{8} $ and $ 8x $ to get $ \dfrac{ \color{purple}{ -12x^3+96x^2+64x } }{ 8 }$.

Step 1: Write $ 8x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{8}$.

$$ \begin{aligned} \frac{-12x^3+96x^2}{8} +8x & \xlongequal{\text{Step 1}} \frac{-12x^3+96x^2}{8} + \frac{8x}{\color{red}{1}} = \frac{ -12x^3+96x^2 }{ 8 } + \frac{ 8x \cdot \color{blue}{ 8 }}{ 1 \cdot \color{blue}{ 8 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -12x^3+96x^2 } }{ 8 } + \frac{ \color{purple}{ 64x } }{ 8 }=\frac{ \color{purple}{ -12x^3+96x^2+64x } }{ 8 } \end{aligned} $$
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