Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org

◀ back to index

Question

Answer

$$(\dfrac{2}{7}x^3+\dfrac{1}{2}xy^2-\dfrac{1}{5}x^2y)(\dfrac{1}{4}x^2-\dfrac{2}{3}xy+\dfrac{5}{6}y^2)=\dfrac{60x^5-202x^4y+417x^3y^2-420x^2y^3+350xy^4}{840}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}(\frac{2}{7}x^3+\frac{1}{2}xy^2-\frac{1}{5}x^2y)(\frac{1}{4}x^2-\frac{2}{3}xy+\frac{5}{6}y^2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}(\frac{2x^3}{7}+\frac{x}{2}y^2-\frac{x^2}{5}y)(\frac{x^2}{4}-\frac{2x}{3}y+\frac{5y^2}{6}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}(\frac{2x^3}{7}+\frac{xy^2}{2}-\frac{x^2y}{5})(\frac{x^2}{4}-\frac{2xy}{3}+\frac{5y^2}{6}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle13}{\textcircled {13}} \htmlClass{explanationCircle explanationCircle14}{\textcircled {14}} \htmlClass{explanationCircle explanationCircle15}{\textcircled {15}} \htmlClass{explanationCircle explanationCircle16}{\textcircled {16}} } }}}(\frac{4x^3+7xy^2}{14}-\frac{x^2y}{5})(\frac{3x^2-8xy}{12}+\frac{5y^2}{6}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle17}{\textcircled {17}} \htmlClass{explanationCircle explanationCircle18}{\textcircled {18}} } }}}\frac{20x^3-14x^2y+35xy^2}{70}\frac{3x^2-8xy+10y^2}{12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle19}{\textcircled {19}} } }}}\frac{60x^5-202x^4y+417x^3y^2-420x^2y^3+350xy^4}{840}\end{aligned} $$
①

Multiply $ \dfrac{2}{7} $ by $ x^3 $ to get $ \dfrac{ 2x^3 }{ 7 } $.

Step 1: Write $ x^3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2}{7} \cdot x^3 & \xlongequal{\text{Step 1}} \frac{2}{7} \cdot \frac{x^3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x^3 }{ 7 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^3 }{ 7 } \end{aligned} $$
②

Multiply $ \dfrac{1}{2} $ by $ x $ to get $ \dfrac{ x }{ 2 } $.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{2} \cdot x & \xlongequal{\text{Step 1}} \frac{1}{2} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x }{ 2 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x }{ 2 } \end{aligned} $$
③

Multiply $ \dfrac{1}{5} $ by $ x^2 $ to get $ \dfrac{ x^2 }{ 5 } $.

Step 1: Write $ x^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Multiply $ \dfrac{1}{4} $ by $ x^2 $ to get $ \dfrac{ x^2 }{ 4 } $.

Step 1: Write $ x^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Multiply $ \dfrac{2}{3} $ by $ x $ to get $ \dfrac{ 2x }{ 3 } $.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{2}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 2x }{ 3 } \end{aligned} $$
⑥

Multiply $ \dfrac{5}{6} $ by $ y^2 $ to get $ \dfrac{ 5y^2 }{ 6 } $.

Step 1: Write $ y^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{5}{6} \cdot y^2 & \xlongequal{\text{Step 1}} \frac{5}{6} \cdot \frac{y^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 5 \cdot y^2 }{ 6 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5y^2 }{ 6 } \end{aligned} $$
⑦

Multiply $ \dfrac{2}{7} $ by $ x^3 $ to get $ \dfrac{ 2x^3 }{ 7 } $.

Step 1: Write $ x^3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2}{7} \cdot x^3 & \xlongequal{\text{Step 1}} \frac{2}{7} \cdot \frac{x^3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x^3 }{ 7 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^3 }{ 7 } \end{aligned} $$
⑧

Multiply $ \dfrac{x}{2} $ by $ y^2 $ to get $ \dfrac{ xy^2 }{ 2 } $.

Step 1: Write $ y^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x}{2} \cdot y^2 & \xlongequal{\text{Step 1}} \frac{x}{2} \cdot \frac{y^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ x \cdot y^2 }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ xy^2 }{ 2 } \end{aligned} $$
⑨

Multiply $ \dfrac{x^2}{5} $ by $ y $ to get $ \dfrac{ x^2y }{ 5 } $.

Step 1: Write $ y $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^2}{5} \cdot y & \xlongequal{\text{Step 1}} \frac{x^2}{5} \cdot \frac{y}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ x^2 \cdot y }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2y }{ 5 } \end{aligned} $$
⑩

Multiply $ \dfrac{1}{4} $ by $ x^2 $ to get $ \dfrac{ x^2 }{ 4 } $.

Step 1: Write $ x^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{4} \cdot x^2 & \xlongequal{\text{Step 1}} \frac{1}{4} \cdot \frac{x^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x^2 }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2 }{ 4 } \end{aligned} $$
⑪

Multiply $ \dfrac{2x}{3} $ by $ y $ to get $ \dfrac{ 2xy }{ 3 } $.

Step 1: Write $ y $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2x}{3} \cdot y & \xlongequal{\text{Step 1}} \frac{2x}{3} \cdot \frac{y}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2x \cdot y }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2xy }{ 3 } \end{aligned} $$
⑫

Multiply $ \dfrac{5}{6} $ by $ y^2 $ to get $ \dfrac{ 5y^2 }{ 6 } $.

Step 1: Write $ y^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{5}{6} \cdot y^2 & \xlongequal{\text{Step 1}} \frac{5}{6} \cdot \frac{y^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 5 \cdot y^2 }{ 6 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5y^2 }{ 6 } \end{aligned} $$
⑬

Add $ \dfrac{2x^3}{7} $ and $ \dfrac{xy^2}{2} $ to get $ \dfrac{ \color{purple}{ 4x^3+7xy^2 } }{ 14 }$.

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}{ 2 }$ and the second by $\color{blue}{ 7 }$.

$$ \begin{aligned} \frac{2x^3}{7} + \frac{xy^2}{2} & = \frac{ 2x^3 \cdot \color{blue}{ 2 }}{ 7 \cdot \color{blue}{ 2 }} + \frac{ xy^2 \cdot \color{blue}{ 7 }}{ 2 \cdot \color{blue}{ 7 }} = \\[1ex] &=\frac{ \color{purple}{ 4x^3 } }{ 14 } + \frac{ \color{purple}{ 7xy^2 } }{ 14 }=\frac{ \color{purple}{ 4x^3+7xy^2 } }{ 14 } \end{aligned} $$
⑭

Multiply $ \dfrac{x^2}{5} $ by $ y $ to get $ \dfrac{ x^2y }{ 5 } $.

Step 1: Write $ y $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^2}{5} \cdot y & \xlongequal{\text{Step 1}} \frac{x^2}{5} \cdot \frac{y}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ x^2 \cdot y }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2y }{ 5 } \end{aligned} $$
⑮

Subtract $ \dfrac{2xy}{3} $ from $ \dfrac{x^2}{4} $ to get $ \dfrac{ \color{purple}{ 3x^2-8xy } }{ 12 }$.

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}{ 3 }$ and the second by $\color{blue}{ 4 }$.

$$ \begin{aligned} \frac{x^2}{4} - \frac{2xy}{3} & = \frac{ x^2 \cdot \color{blue}{ 3 }}{ 4 \cdot \color{blue}{ 3 }} - \frac{ 2xy \cdot \color{blue}{ 4 }}{ 3 \cdot \color{blue}{ 4 }} = \\[1ex] &=\frac{ \color{purple}{ 3x^2 } }{ 12 } - \frac{ \color{purple}{ 8xy } }{ 12 }=\frac{ \color{purple}{ 3x^2-8xy } }{ 12 } \end{aligned} $$
⑯

Multiply $ \dfrac{5}{6} $ by $ y^2 $ to get $ \dfrac{ 5y^2 }{ 6 } $.

Step 1: Write $ y^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{5}{6} \cdot y^2 & \xlongequal{\text{Step 1}} \frac{5}{6} \cdot \frac{y^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 5 \cdot y^2 }{ 6 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5y^2 }{ 6 } \end{aligned} $$
⑰

Subtract $ \dfrac{x^2y}{5} $ from $ \dfrac{4x^3+7xy^2}{14} $ to get $ \dfrac{ \color{purple}{ 20x^3-14x^2y+35xy^2 } }{ 70 }$.

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}{ 5 }$ and the second by $\color{blue}{ 14 }$.

$$ \begin{aligned} \frac{4x^3+7xy^2}{14} - \frac{x^2y}{5} & = \frac{ \left( 4x^3+7xy^2 \right) \cdot \color{blue}{ 5 }}{ 14 \cdot \color{blue}{ 5 }} - \frac{ x^2y \cdot \color{blue}{ 14 }}{ 5 \cdot \color{blue}{ 14 }} = \\[1ex] &=\frac{ \color{purple}{ 20x^3+35xy^2 } }{ 70 } - \frac{ \color{purple}{ 14x^2y } }{ 70 }=\frac{ \color{purple}{ 20x^3-14x^2y+35xy^2 } }{ 70 } \end{aligned} $$
⑱

Add $ \dfrac{3x^2-8xy}{12} $ and $ \dfrac{5y^2}{6} $ to get $ \dfrac{ \color{purple}{ 3x^2-8xy+10y^2 } }{ 12 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{2}$.

$$ \begin{aligned} \frac{3x^2-8xy}{12} + \frac{5y^2}{6} & = \frac{ 3x^2-8xy }{ 12 } + \frac{ 5y^2 \cdot \color{blue}{ 2 }}{ 6 \cdot \color{blue}{ 2 }} = \\[1ex] &=\frac{ \color{purple}{ 3x^2-8xy } }{ 12 } + \frac{ \color{purple}{ 10y^2 } }{ 12 }=\frac{ \color{purple}{ 3x^2-8xy+10y^2 } }{ 12 } \end{aligned} $$
⑲

Multiply $ \dfrac{20x^3-14x^2y+35xy^2}{70} $ by $ \dfrac{3x^2-8xy+10y^2}{12} $ to get $ \dfrac{60x^5-202x^4y+417x^3y^2-420x^2y^3+350xy^4}{840} $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{20x^3-14x^2y+35xy^2}{70} \cdot \frac{3x^2-8xy+10y^2}{12} & \xlongequal{\text{Step 1}} \frac{ \left( 20x^3-14x^2y+35xy^2 \right) \cdot \left( 3x^2-8xy+10y^2 \right) }{ 70 \cdot 12 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 60x^5-160x^4y+200x^3y^2-42x^4y+112x^3y^2-140x^2y^3+105x^3y^2-280x^2y^3+350xy^4 }{ 840 } = \frac{60x^5-202x^4y+417x^3y^2-420x^2y^3+350xy^4}{840} \end{aligned} $$
This page was created using
Simplify polynomials calculator