Question
Answer
$$(1-t)\cdot(9-t)(4-t)^2=t^4-18t^3+105t^2-232t+144$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-t)\cdot(9-t)(4-t)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1-t)\cdot(9-t)(16-8t+t^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(9-t-9t+t^2)(16-8t+t^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(1t^2-10t+9)(16-8t+t^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}t^4-18t^3+105t^2-232t+144\end{aligned} $$ | |
| 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( 9-t\right) $. $$ \left( \color{blue}{1-t}\right) \cdot \left( 9-t\right) = 9-t-9t+t^2 $$ |
| ③ | Combine like terms: $$ 9 \color{blue}{-t} \color{blue}{-9t} +t^2 = t^2 \color{blue}{-10t} +9 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{t^2-10t+9}\right) $ by each term in $ \left( 16-8t+t^2\right) $. $$ \left( \color{blue}{t^2-10t+9}\right) \cdot \left( 16-8t+t^2\right) = 16t^2-8t^3+t^4-160t+80t^2-10t^3+144-72t+9t^2 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{16t^2} \color{red}{-8t^3} +t^4 \color{green}{-160t} + \color{orange}{80t^2} \color{red}{-10t^3} +144 \color{green}{-72t} + \color{orange}{9t^2} = \\ = t^4 \color{red}{-18t^3} + \color{orange}{105t^2} \color{green}{-232t} +144 $$ |
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