Answer
$$(2x-1)(3x^2-3x-1)=6x^3-9x^2+x+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x-1)(3x^2-3x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6x^3-6x^2-2x-3x^2+3x+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^3-9x^2+x+1\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x-1}\right) $ by each term in $ \left( 3x^2-3x-1\right) $. $$ \left( \color{blue}{2x-1}\right) \cdot \left( 3x^2-3x-1\right) = 6x^3-6x^2-2x-3x^2+3x+1 $$ |
| ② | Combine like terms: $$ 6x^3 \color{blue}{-6x^2} \color{red}{-2x} \color{blue}{-3x^2} + \color{red}{3x} +1 = 6x^3 \color{blue}{-9x^2} + \color{red}{x} +1 $$ |
This page was created using
Simplify polynomials calculator