Question
Answer
$$(2x^2+x-2)(-2x^2-3x+8)=-4x^4-8x^3+17x^2+14x-16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x^2+x-2)(-2x^2-3x+8)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-4x^4-8x^3+17x^2+14x-16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x^2+x-2}\right) $ by each term in $ \left( -2x^2-3x+8\right) $. $$ \left( \color{blue}{2x^2+x-2}\right) \cdot \left( -2x^2-3x+8\right) = -4x^4-6x^3+16x^2-2x^3-3x^2+8x+4x^2+6x-16 $$ |
| ② | Combine like terms: $$ -4x^4 \color{blue}{-6x^3} + \color{red}{16x^2} \color{blue}{-2x^3} \color{green}{-3x^2} + \color{orange}{8x} + \color{green}{4x^2} + \color{orange}{6x} -16 = \\ = -4x^4 \color{blue}{-8x^3} + \color{green}{17x^2} + \color{orange}{14x} -16 $$ |
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