Question
Answer
$$(x-3)^2(x+2)=x^3-4x^2-3x+18$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-3)^2(x+2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^2-6x+9)(x+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^3+2x^2-6x^2-12x+9x+18 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^3-4x^2-3x+18\end{aligned} $$ | |
| 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 each term of $ \left( \color{blue}{x^2-6x+9}\right) $ by each term in $ \left( x+2\right) $. $$ \left( \color{blue}{x^2-6x+9}\right) \cdot \left( x+2\right) = x^3+2x^2-6x^2-12x+9x+18 $$ |
| ③ | Combine like terms: $$ x^3+ \color{blue}{2x^2} \color{blue}{-6x^2} \color{red}{-12x} + \color{red}{9x} +18 = x^3 \color{blue}{-4x^2} \color{red}{-3x} +18 $$ |
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