Question
Answer
$$-(s^2+2s)(s^2+8s+4)-(s^2+2s)\cdot5s=-s^4-15s^3-30s^2-8s$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-(s^2+2s)(s^2+8s+4)-(s^2+2s)\cdot5s& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(1s^4+8s^3+4s^2+2s^3+16s^2+8s)-(5s^3+10s^2) \xlongequal{ } \\[1 em] & \xlongequal{ }-(1s^4+10s^3+20s^2+8s)-(5s^3+10s^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-s^4-10s^3-20s^2-8s-(5s^3+10s^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-s^4-10s^3-20s^2-8s-5s^3-10s^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-s^4-15s^3-30s^2-8s\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{s^2+2s}\right) $ by each term in $ \left( s^2+8s+4\right) $. $$ \left( \color{blue}{s^2+2s}\right) \cdot \left( s^2+8s+4\right) = s^4+8s^3+4s^2+2s^3+16s^2+8s $$$$ \left( \color{blue}{s^2+2s}\right) \cdot 5s = 5s^3+10s^2 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(s^4+10s^3+20s^2+8s \right) = -s^4-10s^3-20s^2-8s $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 5s^3+10s^2 \right) = -5s^3-10s^2 $$ |
| ④ | Combine like terms: $$ -s^4 \color{blue}{-10s^3} \color{red}{-20s^2} -8s \color{blue}{-5s^3} \color{red}{-10s^2} = -s^4 \color{blue}{-15s^3} \color{red}{-30s^2} -8s $$ |
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