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Question

Answer

$$-\dfrac{x^2}{8}+1+5x^3-x^4=\dfrac{-8x^4+40x^3-x^2+8}{8}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}-\frac{x^2}{8}+1+5x^3-x^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-x^2+8}{8}+5x^3-x^4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{40x^3-x^2+8}{8}-x^4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-8x^4+40x^3-x^2+8}{8}\end{aligned} $$
①

Add $ \dfrac{-x^2}{8} $ and $ 1 $ to get $ \dfrac{ \color{purple}{ -x^2+8 } }{ 8 }$.

Step 1: Write $ 1 $ 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 $\color{blue}{8}$.

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

Add $ \dfrac{-x^2+8}{8} $ and $ 5x^3 $ to get $ \dfrac{ \color{purple}{ 40x^3-x^2+8 } }{ 8 }$.

Step 1: Write $ 5x^3 $ 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 $\color{blue}{8}$.

$$ \begin{aligned} \frac{-x^2+8}{8} +5x^3 & \xlongequal{\text{Step 1}} \frac{-x^2+8}{8} + \frac{5x^3}{\color{red}{1}} = \frac{ -x^2+8 }{ 8 } + \frac{ 5x^3 \cdot \color{blue}{ 8 }}{ 1 \cdot \color{blue}{ 8 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -x^2+8 } }{ 8 } + \frac{ \color{purple}{ 40x^3 } }{ 8 }=\frac{ \color{purple}{ 40x^3-x^2+8 } }{ 8 } \end{aligned} $$
③

Subtract $x^4$ from $ \dfrac{40x^3-x^2+8}{8} $ to get $ \dfrac{ \color{purple}{ -8x^4+40x^3-x^2+8 } }{ 8 }$.

Step 1: Write $ x^4 $ 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 second fraction by $\color{blue}{8}$.

$$ \begin{aligned} \frac{40x^3-x^2+8}{8} -x^4 & \xlongequal{\text{Step 1}} \frac{40x^3-x^2+8}{8} - \frac{x^4}{\color{red}{1}} = \frac{ 40x^3-x^2+8 }{ 8 } - \frac{ x^4 \cdot \color{blue}{ 8 }}{ 1 \cdot \color{blue}{ 8 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 40x^3-x^2+8 } }{ 8 } - \frac{ \color{purple}{ 8x^4 } }{ 8 }=\frac{ \color{purple}{ -8x^4+40x^3-x^2+8 } }{ 8 } \end{aligned} $$
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