Question
Answer
$$(1-t)\cdot(5-t)(4-t)^2=t^4-14t^3+69t^2-136t+80$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-t)\cdot(5-t)(4-t)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1-t)\cdot(5-t)(16-8t+t^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(5-t-5t+t^2)(16-8t+t^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(1t^2-6t+5)(16-8t+t^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}t^4-14t^3+69t^2-136t+80\end{aligned} $$ | |
| ① | Find $ \left(4-t\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 4 } $ and $ B = \color{red}{ t }$. $$ \begin{aligned}\left(4-t\right)^2 = \color{blue}{4^2} -2 \cdot 4 \cdot t + \color{red}{t^2} = 16-8t+t^2\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{1-t}\right) $ by each term in $ \left( 5-t\right) $. $$ \left( \color{blue}{1-t}\right) \cdot \left( 5-t\right) = 5-t-5t+t^2 $$ |
| ③ | Combine like terms: $$ 5 \color{blue}{-t} \color{blue}{-5t} +t^2 = t^2 \color{blue}{-6t} +5 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{t^2-6t+5}\right) $ by each term in $ \left( 16-8t+t^2\right) $. $$ \left( \color{blue}{t^2-6t+5}\right) \cdot \left( 16-8t+t^2\right) = 16t^2-8t^3+t^4-96t+48t^2-6t^3+80-40t+5t^2 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{16t^2} \color{red}{-8t^3} +t^4 \color{green}{-96t} + \color{orange}{48t^2} \color{red}{-6t^3} +80 \color{green}{-40t} + \color{orange}{5t^2} = \\ = t^4 \color{red}{-14t^3} + \color{orange}{69t^2} \color{green}{-136t} +80 $$ |
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