Question
Answer
$$0.4(t-4)^3-10(t-4)=-10t+40$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}0.4(t-4)^3-10(t-4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}0.4(1t^3-12t^2+48t-64)-10(t-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}0t^3+0t^2+0t+0-(10t-40) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}0t^3+0t^2+0t+0-10t+40 \xlongequal{ } \\[1 em] & \xlongequal{ }0t^30t^20t0-10t+40 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-10t+40\end{aligned} $$ | |
| ① | Find $ \left(t-4\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = t $ and $ B = 4 $. $$ \left(t-4\right)^3 = t^3-3 \cdot t^2 \cdot 4 + 3 \cdot t \cdot 4^2-4^3 = t^3-12t^2+48t-64 $$ |
| ② | Multiply $ \color{blue}{0} $ by $ \left( t^3-12t^2+48t-64\right) $ $$ \color{blue}{0} \cdot \left( t^3-12t^2+48t-64\right) = 0t^30t^20t0 $$Multiply $ \color{blue}{10} $ by $ \left( t-4\right) $ $$ \color{blue}{10} \cdot \left( t-4\right) = 10t-40 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 10t-40 \right) = -10t+40 $$ |
| ④ | Combine like terms: $$ 0t^30t^2 \color{blue}{0t} \color{red}{0} \color{blue}{-10t} + \color{red}{40} = \color{blue}{-10t} + \color{red}{40} $$ |
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