Question
Answer
$$(3x^2-16x-12)(x-6)=3x^3-34x^2+84x+72$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x^2-16x-12)(x-6)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3x^3-18x^2-16x^2+96x-12x+72 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^3-34x^2+84x+72\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x^2-16x-12}\right) $ by each term in $ \left( x-6\right) $. $$ \left( \color{blue}{3x^2-16x-12}\right) \cdot \left( x-6\right) = 3x^3-18x^2-16x^2+96x-12x+72 $$ |
| ② | Combine like terms: $$ 3x^3 \color{blue}{-18x^2} \color{blue}{-16x^2} + \color{red}{96x} \color{red}{-12x} +72 = 3x^3 \color{blue}{-34x^2} + \color{red}{84x} +72 $$ |
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