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