Question
Answer
$$(3x^3-4x+8)(4x-8)=12x^4-24x^3-16x^2+64x-64$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x^3-4x+8)(4x-8)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}12x^4-24x^3-16x^2+32x+32x-64 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}12x^4-24x^3-16x^2+64x-64\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x^3-4x+8}\right) $ by each term in $ \left( 4x-8\right) $. $$ \left( \color{blue}{3x^3-4x+8}\right) \cdot \left( 4x-8\right) = 12x^4-24x^3-16x^2+32x+32x-64 $$ |
| ② | Combine like terms: $$ 12x^4-24x^3-16x^2+ \color{blue}{32x} + \color{blue}{32x} -64 = 12x^4-24x^3-16x^2+ \color{blue}{64x} -64 $$ |
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