Answer
$$(-x)^2\cdot(-3-i)=-ix^2-3x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-x)^2\cdot(-3-i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2\cdot(-3-i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-3x^2-ix^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-ix^2-3x^2\end{aligned} $$ | |
| ① | $$ \left( -x \right)^2 = (-1)^2x^2 = x^2 $$ |
| ② | Multiply $ \color{blue}{x^2} $ by $ \left( -3-i\right) $ $$ \color{blue}{x^2} \cdot \left( -3-i\right) = -3x^2-ix^2 $$ |
| ③ | Combine like terms: $$ -ix^2-3x^2 = -ix^2-3x^2 $$ |
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