Question
Answer
$$x(x-1)(x-2)(x^2-5x+7)=x^5-8x^4+24x^3-31x^2+14x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}x(x-1)(x-2)(x^2-5x+7)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^2-x)(x-2)(x^2-5x+7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^3-2x^2-x^2+2x)(x^2-5x+7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(x^3-3x^2+2x)(x^2-5x+7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^5-8x^4+24x^3-31x^2+14x\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{x} $ by $ \left( x-1\right) $ $$ \color{blue}{x} \cdot \left( x-1\right) = x^2-x $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^2-x}\right) $ by each term in $ \left( x-2\right) $. $$ \left( \color{blue}{x^2-x}\right) \cdot \left( x-2\right) = x^3-2x^2-x^2+2x $$ |
| ③ | Combine like terms: $$ x^3 \color{blue}{-2x^2} \color{blue}{-x^2} +2x = x^3 \color{blue}{-3x^2} +2x $$ |
| ④ | Multiply each term of $ \left( \color{blue}{x^3-3x^2+2x}\right) $ by each term in $ \left( x^2-5x+7\right) $. $$ \left( \color{blue}{x^3-3x^2+2x}\right) \cdot \left( x^2-5x+7\right) = x^5-5x^4+7x^3-3x^4+15x^3-21x^2+2x^3-10x^2+14x $$ |
| ⑤ | Combine like terms: $$ x^5 \color{blue}{-5x^4} + \color{red}{7x^3} \color{blue}{-3x^4} + \color{green}{15x^3} \color{orange}{-21x^2} + \color{green}{2x^3} \color{orange}{-10x^2} +14x = \\ = x^5 \color{blue}{-8x^4} + \color{green}{24x^3} \color{orange}{-31x^2} +14x $$ |
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