Answer
$$(x-6)(x+1)^2=x^3-4x^2-11x-6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-6)(x+1)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x-6)(x^2+2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^3+2x^2+x-6x^2-12x-6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^3-4x^2-11x-6\end{aligned} $$ | |
| ① | Find $ \left(x+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}{ x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(x+1\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 1 + \color{red}{1^2} = x^2+2x+1\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x-6}\right) $ by each term in $ \left( x^2+2x+1\right) $. $$ \left( \color{blue}{x-6}\right) \cdot \left( x^2+2x+1\right) = x^3+2x^2+x-6x^2-12x-6 $$ |
| ③ | Combine like terms: $$ x^3+ \color{blue}{2x^2} + \color{red}{x} \color{blue}{-6x^2} \color{red}{-12x} -6 = x^3 \color{blue}{-4x^2} \color{red}{-11x} -6 $$ |
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