Answer
$$(x-4)(3x^2-6x+4)=3x^3-18x^2+28x-16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-4)(3x^2-6x+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3x^3-6x^2+4x-12x^2+24x-16 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^3-18x^2+28x-16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x-4}\right) $ by each term in $ \left( 3x^2-6x+4\right) $. $$ \left( \color{blue}{x-4}\right) \cdot \left( 3x^2-6x+4\right) = 3x^3-6x^2+4x-12x^2+24x-16 $$ |
| ② | Combine like terms: $$ 3x^3 \color{blue}{-6x^2} + \color{red}{4x} \color{blue}{-12x^2} + \color{red}{24x} -16 = 3x^3 \color{blue}{-18x^2} + \color{red}{28x} -16 $$ |
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