Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org

◀ back to index

Question

Answer

$$x-\dfrac{3}{x^2}+4x-45=\dfrac{5x^3-45x^2-3}{x^2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}x-\frac{3}{x^2}+4x-45& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^3-3}{x^2}+4x-45 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5x^3-3}{x^2}-45 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5x^3-45x^2-3}{x^2}\end{aligned} $$
①

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

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

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

Add $ \dfrac{x^3-3}{x^2} $ and $ 4x $ to get $ \dfrac{ \color{purple}{ 5x^3-3 } }{ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{x^3-3}{x^2} +4x & \xlongequal{\text{Step 1}} \frac{x^3-3}{x^2} + \frac{4x}{\color{red}{1}} = \frac{ x^3-3 }{ x^2 } + \frac{ 4x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^3-3 } }{ x^2 } + \frac{ \color{purple}{ 4x^3 } }{ x^2 }=\frac{ \color{purple}{ 5x^3-3 } }{ x^2 } \end{aligned} $$
③

Subtract $45$ from $ \dfrac{5x^3-3}{x^2} $ to get $ \dfrac{ \color{purple}{ 5x^3-45x^2-3 } }{ x^2 }$.

Step 1: Write $ 45 $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{5x^3-3}{x^2} -45 & \xlongequal{\text{Step 1}} \frac{5x^3-3}{x^2} - \frac{45}{\color{red}{1}} = \frac{ 5x^3-3 }{ x^2 } - \frac{ 45 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 5x^3-3 } }{ x^2 } - \frac{ \color{purple}{ 45x^2 } }{ x^2 }=\frac{ \color{purple}{ 5x^3-45x^2-3 } }{ x^2 } \end{aligned} $$
This page was created using
Simplify Rational Expressions