Question
Answer
$$-(x+4)^2(x-1)^2(x+2)(3x-2)=-3x^6-22x^5-23x^4+92x^3+52x^2-160x+64$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-(x+4)^2(x-1)^2(x+2)(3x-2)& \xlongequal{ }-(x^2+8x+16)(x^2-2x+1)(x+2)(3x-2) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^4+6x^3+x^2-24x+16)(x+2)(3x-2) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^5+8x^4+13x^3-22x^2-32x+32)(3x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-(3x^6+22x^5+23x^4-92x^3-52x^2+160x-64) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-3x^6-22x^5-23x^4+92x^3+52x^2-160x+64\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^5+8x^4+13x^3-22x^2-32x+32}\right) $ by each term in $ \left( 3x-2\right) $. $$ \left( \color{blue}{x^5+8x^4+13x^3-22x^2-32x+32}\right) \cdot \left( 3x-2\right) = \\ = 3x^6-2x^5+24x^5-16x^4+39x^4-26x^3-66x^3+44x^2-96x^2+64x+96x-64 $$ |
| ② | Combine like terms: $$ 3x^6 \color{blue}{-2x^5} + \color{blue}{24x^5} \color{red}{-16x^4} + \color{red}{39x^4} \color{green}{-26x^3} \color{green}{-66x^3} + \color{orange}{44x^2} \color{orange}{-96x^2} + \color{blue}{64x} + \color{blue}{96x} -64 = \\ = 3x^6+ \color{blue}{22x^5} + \color{red}{23x^4} \color{green}{-92x^3} \color{orange}{-52x^2} + \color{blue}{160x} -64 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left(3x^6+22x^5+23x^4-92x^3-52x^2+160x-64 \right) = -3x^6-22x^5-23x^4+92x^3+52x^2-160x+64 $$ |
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