Question
Answer
$$(x^2-3x-3)(2x^2+6x-5)=2x^4-29x^2-3x+15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2-3x-3)(2x^2+6x-5)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2x^4-29x^2-3x+15\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2-3x-3}\right) $ by each term in $ \left( 2x^2+6x-5\right) $. $$ \left( \color{blue}{x^2-3x-3}\right) \cdot \left( 2x^2+6x-5\right) = \\ = 2x^4+ \cancel{6x^3}-5x^2 -\cancel{6x^3}-18x^2+15x-6x^2-18x+15 $$ |
| ② | Combine like terms: $$ 2x^4+ \, \color{blue}{ \cancel{6x^3}} \, \color{green}{-5x^2} \, \color{blue}{ -\cancel{6x^3}} \, \color{orange}{-18x^2} + \color{blue}{15x} \color{orange}{-6x^2} \color{blue}{-18x} +15 = 2x^4 \color{orange}{-29x^2} \color{blue}{-3x} +15 $$ |
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