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Question

Answer

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

Explanation

Tap the blue circles to see an explanation.

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

Subtract $ \dfrac{1}{x^2} $ from $ x^2 $ to get $ \dfrac{ \color{purple}{ x^4-1 } }{ x^2 }$.

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

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

Add $ \dfrac{x^4-1}{x^2} $ and $ 2x $ to get $ \dfrac{ \color{purple}{ x^4+2x^3-1 } }{ x^2 }$.

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

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

Add $ \dfrac{x^4+2x^3-1}{x^2} $ and $ 1 $ to get $ \dfrac{ \color{purple}{ x^4+2x^3+x^2-1 } }{ x^2 }$.

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 $\color{blue}{x^2}$.

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