Question
Answer
$$(x^2+16)(x+2)(2x^2-5x-3)=2x^5-x^4+19x^3-22x^2-208x-96$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x^2+16)(x+2)(2x^2-5x-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^3+2x^2+16x+32)(2x^2-5x-3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2x^5-x^4+19x^3-22x^2-208x-96\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x^2+16}\right) $ by each term in $ \left( x+2\right) $. $$ \left( \color{blue}{x^2+16}\right) \cdot \left( x+2\right) = x^3+2x^2+16x+32 $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^3+2x^2+16x+32}\right) $ by each term in $ \left( 2x^2-5x-3\right) $. $$ \left( \color{blue}{x^3+2x^2+16x+32}\right) \cdot \left( 2x^2-5x-3\right) = \\ = 2x^5-5x^4-3x^3+4x^4-10x^3-6x^2+32x^3-80x^2-48x+64x^2-160x-96 $$ |
| ③ | Combine like terms: $$ 2x^5 \color{blue}{-5x^4} \color{red}{-3x^3} + \color{blue}{4x^4} \color{green}{-10x^3} \color{orange}{-6x^2} + \color{green}{32x^3} \color{blue}{-80x^2} \color{red}{-48x} + \color{blue}{64x^2} \color{red}{-160x} -96 = \\ = 2x^5 \color{blue}{-x^4} + \color{green}{19x^3} \color{blue}{-22x^2} \color{red}{-208x} -96 $$ |
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