Question
Answer
$$x(x-3)^2(x-8)=x^4-14x^3+57x^2-72x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}x(x-3)^2(x-8)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x(x^2-6x+9)(x-8) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^3-6x^2+9x)(x-8) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^4-8x^3-6x^3+48x^2+9x^2-72x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^4-14x^3+57x^2-72x\end{aligned} $$ | |
| ① | Find $ \left(x-3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(x-3\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 3 + \color{red}{3^2} = x^2-6x+9\end{aligned} $$ |
| ② | Multiply $ \color{blue}{x} $ by $ \left( x^2-6x+9\right) $ $$ \color{blue}{x} \cdot \left( x^2-6x+9\right) = x^3-6x^2+9x $$ |
| ③ | Multiply each term of $ \left( \color{blue}{x^3-6x^2+9x}\right) $ by each term in $ \left( x-8\right) $. $$ \left( \color{blue}{x^3-6x^2+9x}\right) \cdot \left( x-8\right) = x^4-8x^3-6x^3+48x^2+9x^2-72x $$ |
| ④ | Combine like terms: $$ x^4 \color{blue}{-8x^3} \color{blue}{-6x^3} + \color{red}{48x^2} + \color{red}{9x^2} -72x = x^4 \color{blue}{-14x^3} + \color{red}{57x^2} -72x $$ |
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