Question
Answer
$$(y-2)(2y^2+3y-1)=2y^3-y^2-7y+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(y-2)(2y^2+3y-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2y^3+3y^2-y-4y^2-6y+2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2y^3-y^2-7y+2\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{y-2}\right) $ by each term in $ \left( 2y^2+3y-1\right) $. $$ \left( \color{blue}{y-2}\right) \cdot \left( 2y^2+3y-1\right) = 2y^3+3y^2-y-4y^2-6y+2 $$ |
| ② | Combine like terms: $$ 2y^3+ \color{blue}{3y^2} \color{red}{-y} \color{blue}{-4y^2} \color{red}{-6y} +2 = 2y^3 \color{blue}{-y^2} \color{red}{-7y} +2 $$ |
This page was created using
Simplify polynomials calculator