Question
Answer
$$(3x^2-x+2)(2x-10)=6x^3-32x^2+14x-20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x^2-x+2)(2x-10)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^3-30x^2-2x^2+10x+4x-20 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^3-32x^2+14x-20\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x^2-x+2}\right) $ by each term in $ \left( 2x-10\right) $. $$ \left( \color{blue}{3x^2-x+2}\right) \cdot \left( 2x-10\right) = 6x^3-30x^2-2x^2+10x+4x-20 $$ |
| ② | Combine like terms: $$ 6x^3 \color{blue}{-30x^2} \color{blue}{-2x^2} + \color{red}{10x} + \color{red}{4x} -20 = 6x^3 \color{blue}{-32x^2} + \color{red}{14x} -20 $$ |
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