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Question

Answer

$$x^2+3x-\dfrac{40}{x^2}+9x+8=\dfrac{x^4+12x^3+8x^2-40}{x^2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}x^2+3x-\frac{40}{x^2}+9x+8& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^4+3x^3-40}{x^2}+9x+8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^4+12x^3-40}{x^2}+8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^4+12x^3+8x^2-40}{x^2}\end{aligned} $$
①

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

Step 1: Write $ x^2+3x $ 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 $\color{blue}{ x^2 }$.

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

Add $ \dfrac{x^4+3x^3-40}{x^2} $ and $ 9x $ to get $ \dfrac{ \color{purple}{ x^4+12x^3-40 } }{ x^2 }$.

Step 1: Write $ 9x $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{x^4+3x^3-40}{x^2} +9x & \xlongequal{\text{Step 1}} \frac{x^4+3x^3-40}{x^2} + \frac{9x}{\color{red}{1}} = \frac{ x^4+3x^3-40 }{ x^2 } + \frac{ 9x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+3x^3-40 } }{ x^2 } + \frac{ \color{purple}{ 9x^3 } }{ x^2 }=\frac{ \color{purple}{ x^4+12x^3-40 } }{ x^2 } \end{aligned} $$
③

Add $ \dfrac{x^4+12x^3-40}{x^2} $ and $ 8 $ to get $ \dfrac{ \color{purple}{ x^4+12x^3+8x^2-40 } }{ x^2 }$.

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

$$ \begin{aligned} \frac{x^4+12x^3-40}{x^2} +8 & \xlongequal{\text{Step 1}} \frac{x^4+12x^3-40}{x^2} + \frac{8}{\color{red}{1}} = \frac{ x^4+12x^3-40 }{ x^2 } + \frac{ 8 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+12x^3-40 } }{ x^2 } + \frac{ \color{purple}{ 8x^2 } }{ x^2 }=\frac{ \color{purple}{ x^4+12x^3+8x^2-40 } }{ x^2 } \end{aligned} $$
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