Question
Answer
$$4(v-h)\cdot(1+v^2)+(h+3v)((v+h)^2+4)=h^3+5h^2v+3hv^2+7v^3+16v$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}4(v-h)\cdot(1+v^2)+(h+3v)((v+h)^2+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4(v-h)\cdot(1+v^2)+(h+3v)(1v^2+2hv+h^2+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(4v-4h)\cdot(1+v^2)+h^3+5h^2v+7hv^2+3v^3+4h+12v \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4v+4v^3-4h-4hv^2+h^3+5h^2v+7hv^2+3v^3+4h+12v \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}h^3+5h^2v+3hv^2+7v^3+16v\end{aligned} $$ | |
| ① | Find $ \left(v+h\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ v } $ and $ B = \color{red}{ h }$. $$ \begin{aligned}\left(v+h\right)^2 = \color{blue}{v^2} +2 \cdot v \cdot h + \color{red}{h^2} = v^2+2hv+h^2\end{aligned} $$ |
| ② | Multiply $ \color{blue}{4} $ by $ \left( v-h\right) $ $$ \color{blue}{4} \cdot \left( v-h\right) = 4v-4h $$ Multiply each term of $ \left( \color{blue}{h+3v}\right) $ by each term in $ \left( v^2+2hv+h^2+4\right) $. $$ \left( \color{blue}{h+3v}\right) \cdot \left( v^2+2hv+h^2+4\right) = hv^2+2h^2v+h^3+4h+3v^3+6hv^2+3h^2v+12v $$ |
| ③ | Combine like terms: $$ \color{blue}{hv^2} + \color{red}{2h^2v} +h^3+4h+3v^3+ \color{blue}{6hv^2} + \color{red}{3h^2v} +12v = h^3+ \color{red}{5h^2v} + \color{blue}{7hv^2} +3v^3+4h+12v $$ |
| ④ | Multiply each term of $ \left( \color{blue}{4v-4h}\right) $ by each term in $ \left( 1+v^2\right) $. $$ \left( \color{blue}{4v-4h}\right) \cdot \left( 1+v^2\right) = 4v+4v^3-4h-4hv^2 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{4v} + \color{red}{4v^3} \, \color{green}{ -\cancel{4h}} \, \color{blue}{-4hv^2} +h^3+5h^2v+ \color{blue}{7hv^2} + \color{red}{3v^3} + \, \color{green}{ \cancel{4h}} \,+ \color{blue}{12v} = \\ = h^3+5h^2v+ \color{blue}{3hv^2} + \color{red}{7v^3} + \color{blue}{16v} $$ |
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