Question
Answer
$$(x+3)(x+1)(x-4)^2=x^4-4x^3-13x^2+40x+48$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+3)(x+1)(x-4)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x+3)(x+1)(x^2-8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^2+x+3x+3)(x^2-8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(x^2+4x+3)(x^2-8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^4-4x^3-13x^2+40x+48\end{aligned} $$ | |
| ① | Find $ \left(x-4\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}{ 4 }$. $$ \begin{aligned}\left(x-4\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 4 + \color{red}{4^2} = x^2-8x+16\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x+3}\right) $ by each term in $ \left( x+1\right) $. $$ \left( \color{blue}{x+3}\right) \cdot \left( x+1\right) = x^2+x+3x+3 $$ |
| ③ | Combine like terms: $$ x^2+ \color{blue}{x} + \color{blue}{3x} +3 = x^2+ \color{blue}{4x} +3 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{x^2+4x+3}\right) $ by each term in $ \left( x^2-8x+16\right) $. $$ \left( \color{blue}{x^2+4x+3}\right) \cdot \left( x^2-8x+16\right) = x^4-8x^3+16x^2+4x^3-32x^2+64x+3x^2-24x+48 $$ |
| ⑤ | Combine like terms: $$ x^4 \color{blue}{-8x^3} + \color{red}{16x^2} + \color{blue}{4x^3} \color{green}{-32x^2} + \color{orange}{64x} + \color{green}{3x^2} \color{orange}{-24x} +48 = \\ = x^4 \color{blue}{-4x^3} \color{green}{-13x^2} + \color{orange}{40x} +48 $$ |
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