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Question

Answer

$$k^2+13k+\dfrac{40}{k^2}-2k-35=\dfrac{k^4+11k^3-35k^2+40}{k^2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}k^2+13k+\frac{40}{k^2}-2k-35& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{k^4+13k^3+40}{k^2}-2k-35 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{k^4+11k^3+40}{k^2}-35 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{k^4+11k^3-35k^2+40}{k^2}\end{aligned} $$
①

Add $k^2+13k$ and $ \dfrac{40}{k^2} $ to get $ \dfrac{ \color{purple}{ k^4+13k^3+40 } }{ k^2 }$.

Step 1: Write $ k^2+13k $ 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 $\color{blue}{ k^2 }$.

$$ \begin{aligned} k^2+13k+ \frac{40}{k^2} & \xlongequal{\text{Step 1}} \frac{k^2+13k}{\color{red}{1}} + \frac{40}{k^2} = \frac{ \left( k^2+13k \right) \cdot \color{blue}{ k^2 }}{ 1 \cdot \color{blue}{ k^2 }} + \frac{ 40 }{ k^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ k^4+13k^3 } }{ k^2 } + \frac{ \color{purple}{ 40 } }{ k^2 }=\frac{ \color{purple}{ k^4+13k^3+40 } }{ k^2 } \end{aligned} $$
②

Subtract $2k$ from $ \dfrac{k^4+13k^3+40}{k^2} $ to get $ \dfrac{ \color{purple}{ k^4+11k^3+40 } }{ k^2 }$.

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

$$ \begin{aligned} \frac{k^4+13k^3+40}{k^2} -2k & \xlongequal{\text{Step 1}} \frac{k^4+13k^3+40}{k^2} - \frac{2k}{\color{red}{1}} = \frac{ k^4+13k^3+40 }{ k^2 } - \frac{ 2k \cdot \color{blue}{ k^2 }}{ 1 \cdot \color{blue}{ k^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ k^4+13k^3+40 } }{ k^2 } - \frac{ \color{purple}{ 2k^3 } }{ k^2 }=\frac{ \color{purple}{ k^4+11k^3+40 } }{ k^2 } \end{aligned} $$
③

Subtract $35$ from $ \dfrac{k^4+11k^3+40}{k^2} $ to get $ \dfrac{ \color{purple}{ k^4+11k^3-35k^2+40 } }{ k^2 }$.

Step 1: Write $ 35 $ 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 $\color{blue}{k^2}$.

$$ \begin{aligned} \frac{k^4+11k^3+40}{k^2} -35 & \xlongequal{\text{Step 1}} \frac{k^4+11k^3+40}{k^2} - \frac{35}{\color{red}{1}} = \frac{ k^4+11k^3+40 }{ k^2 } - \frac{ 35 \cdot \color{blue}{ k^2 }}{ 1 \cdot \color{blue}{ k^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ k^4+11k^3+40 } }{ k^2 } - \frac{ \color{purple}{ 35k^2 } }{ k^2 }=\frac{ \color{purple}{ k^4+11k^3-35k^2+40 } }{ k^2 } \end{aligned} $$
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