Question
Answer
$$(3x^3-4x+8)(2x^2-8x+4)=6x^5-24x^4+4x^3+48x^2-80x+32$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x^3-4x+8)(2x^2-8x+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^5-24x^4+4x^3+48x^2-80x+32\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x^3-4x+8}\right) $ by each term in $ \left( 2x^2-8x+4\right) $. $$ \left( \color{blue}{3x^3-4x+8}\right) \cdot \left( 2x^2-8x+4\right) = 6x^5-24x^4+12x^3-8x^3+32x^2-16x+16x^2-64x+32 $$ |
| ② | Combine like terms: $$ 6x^5-24x^4+ \color{blue}{12x^3} \color{blue}{-8x^3} + \color{red}{32x^2} \color{green}{-16x} + \color{red}{16x^2} \color{green}{-64x} +32 = \\ = 6x^5-24x^4+ \color{blue}{4x^3} + \color{red}{48x^2} \color{green}{-80x} +32 $$ |
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