Answer
$$-3(x+3)(x-2)(x-8)=-3x^3+21x^2+42x-144$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-3(x+3)(x-2)(x-8)& \xlongequal{ }-(3x+9)(x-2)(x-8) \xlongequal{ } \\[1 em] & \xlongequal{ }-(3x^2-6x+9x-18)(x-8) \xlongequal{ } \\[1 em] & \xlongequal{ }-(3x^2+3x-18)(x-8) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(3x^3-24x^2+3x^2-24x-18x+144) \xlongequal{ } \\[1 em] & \xlongequal{ }-(3x^3-21x^2-42x+144) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-3x^3+21x^2+42x-144\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x^2+3x-18}\right) $ by each term in $ \left( x-8\right) $. $$ \left( \color{blue}{3x^2+3x-18}\right) \cdot \left( x-8\right) = 3x^3-24x^2+3x^2-24x-18x+144 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(3x^3-21x^2-42x+144 \right) = -3x^3+21x^2+42x-144 $$ |
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