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