Answer
$$(2x^3-5x^2+x+2)(3x+2)=6x^4-11x^3-7x^2+8x+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x^3-5x^2+x+2)(3x+2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^4-11x^3-7x^2+8x+4\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x^3-5x^2+x+2}\right) $ by each term in $ \left( 3x+2\right) $. $$ \left( \color{blue}{2x^3-5x^2+x+2}\right) \cdot \left( 3x+2\right) = 6x^4+4x^3-15x^3-10x^2+3x^2+2x+6x+4 $$ |
| ② | Combine like terms: $$ 6x^4+ \color{blue}{4x^3} \color{blue}{-15x^3} \color{red}{-10x^2} + \color{red}{3x^2} + \color{green}{2x} + \color{green}{6x} +4 = 6x^4 \color{blue}{-11x^3} \color{red}{-7x^2} + \color{green}{8x} +4 $$ |
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