Question
Answer
$$-x(x+1)^2(x-3)(x+4)^2=-x^6-7x^5-3x^4+59x^3+104x^2+48x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-x(x+1)^2(x-3)(x+4)^2& \xlongequal{ }-x(x^2+2x+1)(x-3)(x^2+8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^3+2x^2+x)(x-3)(x^2+8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^4-3x^3+2x^3-6x^2+x^2-3x)(x^2+8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^4-x^3-5x^2-3x)(x^2+8x+16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-(x^6+7x^5+3x^4-59x^3-104x^2-48x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-x^6-7x^5-3x^4+59x^3+104x^2+48x\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^4-x^3-5x^2-3x}\right) $ by each term in $ \left( x^2+8x+16\right) $. $$ \left( \color{blue}{x^4-x^3-5x^2-3x}\right) \cdot \left( x^2+8x+16\right) = \\ = x^6+8x^5+16x^4-x^5-8x^4-16x^3-5x^4-40x^3-80x^2-3x^3-24x^2-48x $$ |
| ② | Combine like terms: $$ x^6+ \color{blue}{8x^5} + \color{red}{16x^4} \color{blue}{-x^5} \color{green}{-8x^4} \color{orange}{-16x^3} \color{green}{-5x^4} \color{blue}{-40x^3} \color{red}{-80x^2} \color{blue}{-3x^3} \color{red}{-24x^2} -48x = \\ = x^6+ \color{blue}{7x^5} + \color{green}{3x^4} \color{blue}{-59x^3} \color{red}{-104x^2} -48x $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left(x^6+7x^5+3x^4-59x^3-104x^2-48x \right) = -x^6-7x^5-3x^4+59x^3+104x^2+48x $$ |
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