Answer
$$(2x^2+x-4)(-x^2+3x-2)=-2x^4+5x^3+3x^2-14x+8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x^2+x-4)(-x^2+3x-2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-2x^4+5x^3+3x^2-14x+8\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x^2+x-4}\right) $ by each term in $ \left( -x^2+3x-2\right) $. $$ \left( \color{blue}{2x^2+x-4}\right) \cdot \left( -x^2+3x-2\right) = \\ = -2x^4+6x^3 -\cancel{4x^2}-x^3+3x^2-2x+ \cancel{4x^2}-12x+8 $$ |
| ② | Combine like terms: $$ -2x^4+ \color{blue}{6x^3} \, \color{red}{ -\cancel{4x^2}} \, \color{blue}{-x^3} + \color{orange}{3x^2} \color{blue}{-2x} + \, \color{orange}{ \cancel{4x^2}} \, \color{blue}{-12x} +8 = \\ = -2x^4+ \color{blue}{5x^3} + \color{orange}{3x^2} \color{blue}{-14x} +8 $$ |
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