Answer
$$2s(s-2)^2-14(s-2)^2+1=2s^3-22s^2+64s-55$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2s(s-2)^2-14(s-2)^2+1& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2s(1s^2-4s+4)-14(1s^2-4s+4)+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2s^3-8s^2+8s-(14s^2-56s+56)+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2s^3-8s^2+8s-14s^2+56s-56+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2s^3-22s^2+64s-55\end{aligned} $$ | |
| ① | Find $ \left(s-2\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ s } $ and $ B = \color{red}{ 2 }$. $$ \begin{aligned}\left(s-2\right)^2 = \color{blue}{s^2} -2 \cdot s \cdot 2 + \color{red}{2^2} = s^2-4s+4\end{aligned} $$Find $ \left(s-2\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ s } $ and $ B = \color{red}{ 2 }$. $$ \begin{aligned}\left(s-2\right)^2 = \color{blue}{s^2} -2 \cdot s \cdot 2 + \color{red}{2^2} = s^2-4s+4\end{aligned} $$ |
| ② | Multiply $ \color{blue}{2s} $ by $ \left( s^2-4s+4\right) $ $$ \color{blue}{2s} \cdot \left( s^2-4s+4\right) = 2s^3-8s^2+8s $$Multiply $ \color{blue}{14} $ by $ \left( s^2-4s+4\right) $ $$ \color{blue}{14} \cdot \left( s^2-4s+4\right) = 14s^2-56s+56 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 14s^2-56s+56 \right) = -14s^2+56s-56 $$ |
| ④ | Combine like terms: $$ 2s^3 \color{blue}{-8s^2} + \color{red}{8s} \color{blue}{-14s^2} + \color{red}{56s} \color{green}{-56} + \color{green}{1} = 2s^3 \color{blue}{-22s^2} + \color{red}{64s} \color{green}{-55} $$ |
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