Question
Answer
$$(a-1)(a-2)(a-4)=a^3-7a^2+14a-8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(a-1)(a-2)(a-4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1a^2-2a-a+2)(a-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1a^2-3a+2)(a-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}a^3-4a^2-3a^2+12a+2a-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}a^3-7a^2+14a-8\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{a-1}\right) $ by each term in $ \left( a-2\right) $. $$ \left( \color{blue}{a-1}\right) \cdot \left( a-2\right) = a^2-2a-a+2 $$ |
| ② | Combine like terms: $$ a^2 \color{blue}{-2a} \color{blue}{-a} +2 = a^2 \color{blue}{-3a} +2 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{a^2-3a+2}\right) $ by each term in $ \left( a-4\right) $. $$ \left( \color{blue}{a^2-3a+2}\right) \cdot \left( a-4\right) = a^3-4a^2-3a^2+12a+2a-8 $$ |
| ④ | Combine like terms: $$ a^3 \color{blue}{-4a^2} \color{blue}{-3a^2} + \color{red}{12a} + \color{red}{2a} -8 = a^3 \color{blue}{-7a^2} + \color{red}{14a} -8 $$ |
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