Question
Answer
$$(x-4)(3x^2+3x-1)=3x^3-9x^2-13x+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-4)(3x^2+3x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3x^3+3x^2-x-12x^2-12x+4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^3-9x^2-13x+4\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x-4}\right) $ by each term in $ \left( 3x^2+3x-1\right) $. $$ \left( \color{blue}{x-4}\right) \cdot \left( 3x^2+3x-1\right) = 3x^3+3x^2-x-12x^2-12x+4 $$ |
| ② | Combine like terms: $$ 3x^3+ \color{blue}{3x^2} \color{red}{-x} \color{blue}{-12x^2} \color{red}{-12x} +4 = 3x^3 \color{blue}{-9x^2} \color{red}{-13x} +4 $$ |
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