Question
Answer
$$k(k-1)^2(k-3)=k^4-5k^3+7k^2-3k$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}k(k-1)^2(k-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}k(1k^2-2k+1)(k-3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1k^3-2k^2+k)(k-3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}k^4-3k^3-2k^3+6k^2+k^2-3k \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}k^4-5k^3+7k^2-3k\end{aligned} $$ | |
| ① | Find $ \left(k-1\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ k } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(k-1\right)^2 = \color{blue}{k^2} -2 \cdot k \cdot 1 + \color{red}{1^2} = k^2-2k+1\end{aligned} $$ |
| ② | Multiply $ \color{blue}{k} $ by $ \left( k^2-2k+1\right) $ $$ \color{blue}{k} \cdot \left( k^2-2k+1\right) = k^3-2k^2+k $$ |
| ③ | Multiply each term of $ \left( \color{blue}{k^3-2k^2+k}\right) $ by each term in $ \left( k-3\right) $. $$ \left( \color{blue}{k^3-2k^2+k}\right) \cdot \left( k-3\right) = k^4-3k^3-2k^3+6k^2+k^2-3k $$ |
| ④ | Combine like terms: $$ k^4 \color{blue}{-3k^3} \color{blue}{-2k^3} + \color{red}{6k^2} + \color{red}{k^2} -3k = k^4 \color{blue}{-5k^3} + \color{red}{7k^2} -3k $$ |
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