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Question

Answer

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

Explanation

Tap the blue circles to see an explanation.

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

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

Step 1: Write $ 2x^2 $ 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 $\color{blue}{ x^2 }$.

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

Add $ \dfrac{2x^4-8}{x^2} $ and $ 4x $ to get $ \dfrac{ \color{purple}{ 2x^4+4x^3-8 } }{ x^2 }$.

Step 1: Write $ 4x $ 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 $\color{blue}{x^2}$.

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

Add $ \dfrac{2x^4+4x^3-8}{x^2} $ and $ 4 $ to get $ \dfrac{ \color{purple}{ 2x^4+4x^3+4x^2-8 } }{ x^2 }$.

Step 1: Write $ 4 $ 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 $\color{blue}{x^2}$.

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