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Question

Answer

$$2x^2-26x+\dfrac{72}{x^2}-8x+16=\dfrac{2x^4-34x^3+16x^2+72}{x^2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}2x^2-26x+\frac{72}{x^2}-8x+16& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x^4-26x^3+72}{x^2}-8x+16 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2x^4-34x^3+72}{x^2}+16 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2x^4-34x^3+16x^2+72}{x^2}\end{aligned} $$
①

Add $2x^2-26x$ and $ \dfrac{72}{x^2} $ to get $ \dfrac{ \color{purple}{ 2x^4-26x^3+72 } }{ x^2 }$.

Step 1: Write $ 2x^2-26x $ 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 first fraction by $\color{blue}{ x^2 }$.

$$ \begin{aligned} 2x^2-26x+ \frac{72}{x^2} & \xlongequal{\text{Step 1}} \frac{2x^2-26x}{\color{red}{1}} + \frac{72}{x^2} = \frac{ \left( 2x^2-26x \right) \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} + \frac{ 72 }{ x^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^4-26x^3 } }{ x^2 } + \frac{ \color{purple}{ 72 } }{ x^2 }=\frac{ \color{purple}{ 2x^4-26x^3+72 } }{ x^2 } \end{aligned} $$
②

Subtract $8x$ from $ \dfrac{2x^4-26x^3+72}{x^2} $ to get $ \dfrac{ \color{purple}{ 2x^4-34x^3+72 } }{ x^2 }$.

Step 1: Write $ 8x $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{2x^4-26x^3+72}{x^2} -8x & \xlongequal{\text{Step 1}} \frac{2x^4-26x^3+72}{x^2} - \frac{8x}{\color{red}{1}} = \frac{ 2x^4-26x^3+72 }{ x^2 } - \frac{ 8x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^4-26x^3+72 } }{ x^2 } - \frac{ \color{purple}{ 8x^3 } }{ x^2 }=\frac{ \color{purple}{ 2x^4-34x^3+72 } }{ x^2 } \end{aligned} $$
③

Add $ \dfrac{2x^4-34x^3+72}{x^2} $ and $ 16 $ to get $ \dfrac{ \color{purple}{ 2x^4-34x^3+16x^2+72 } }{ x^2 }$.

Step 1: Write $ 16 $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{2x^4-34x^3+72}{x^2} +16 & \xlongequal{\text{Step 1}} \frac{2x^4-34x^3+72}{x^2} + \frac{16}{\color{red}{1}} = \frac{ 2x^4-34x^3+72 }{ x^2 } + \frac{ 16 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^4-34x^3+72 } }{ x^2 } + \frac{ \color{purple}{ 16x^2 } }{ x^2 }=\frac{ \color{purple}{ 2x^4-34x^3+16x^2+72 } }{ x^2 } \end{aligned} $$
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