Question
Answer
$$x(x-3)(x-2)(x-1)=x^4-6x^3+11x^2-6x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}x(x-3)(x-2)(x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^2-3x)(x-2)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^3-2x^2-3x^2+6x)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(x^3-5x^2+6x)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^4-x^3-5x^3+5x^2+6x^2-6x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^4-6x^3+11x^2-6x\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{x} $ by $ \left( x-3\right) $ $$ \color{blue}{x} \cdot \left( x-3\right) = x^2-3x $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^2-3x}\right) $ by each term in $ \left( x-2\right) $. $$ \left( \color{blue}{x^2-3x}\right) \cdot \left( x-2\right) = x^3-2x^2-3x^2+6x $$ |
| ③ | Combine like terms: $$ x^3 \color{blue}{-2x^2} \color{blue}{-3x^2} +6x = x^3 \color{blue}{-5x^2} +6x $$ |
| ④ | Multiply each term of $ \left( \color{blue}{x^3-5x^2+6x}\right) $ by each term in $ \left( x-1\right) $. $$ \left( \color{blue}{x^3-5x^2+6x}\right) \cdot \left( x-1\right) = x^4-x^3-5x^3+5x^2+6x^2-6x $$ |
| ⑤ | Combine like terms: $$ x^4 \color{blue}{-x^3} \color{blue}{-5x^3} + \color{red}{5x^2} + \color{red}{6x^2} -6x = x^4 \color{blue}{-6x^3} + \color{red}{11x^2} -6x $$ |
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