Answer
$$-(x-5)(x-4)(x-2)=-x^3+11x^2-38x+40$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-(x-5)(x-4)(x-2)& \xlongequal{ }-(x^2-4x-5x+20)(x-2) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^2-9x+20)(x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(x^3-2x^2-9x^2+18x+20x-40) \xlongequal{ } \\[1 em] & \xlongequal{ }-(x^3-11x^2+38x-40) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-x^3+11x^2-38x+40\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2-9x+20}\right) $ by each term in $ \left( x-2\right) $. $$ \left( \color{blue}{x^2-9x+20}\right) \cdot \left( x-2\right) = x^3-2x^2-9x^2+18x+20x-40 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(x^3-11x^2+38x-40 \right) = -x^3+11x^2-38x+40 $$ |
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