Question
Answer
$$-5(3y^2-3y+3)=-15y^2+15y-15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-5(3y^2-3y+3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(15y^2-15y+15) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-15y^2+15y-15\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{5} $ by $ \left( 3y^2-3y+3\right) $ $$ \color{blue}{5} \cdot \left( 3y^2-3y+3\right) = 15y^2-15y+15 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(15y^2-15y+15 \right) = -15y^2+15y-15 $$ |
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