Answer
$$-7x^2(3x^3-4x^2+6x+2)=-21x^5+28x^4-42x^3-14x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-7x^2(3x^3-4x^2+6x+2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(21x^5-28x^4+42x^3+14x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-21x^5+28x^4-42x^3-14x^2\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{7x^2} $ by $ \left( 3x^3-4x^2+6x+2\right) $ $$ \color{blue}{7x^2} \cdot \left( 3x^3-4x^2+6x+2\right) = 21x^5-28x^4+42x^3+14x^2 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(21x^5-28x^4+42x^3+14x^2 \right) = -21x^5+28x^4-42x^3-14x^2 $$ |
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