Question
Answer
$$2x(x^2-7)-4y(x^2+2)+(2y-4)\cdot3x^2=2x^3+2x^2y-12x^2-14x-8y$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2x(x^2-7)-4y(x^2+2)+(2y-4)\cdot3x^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2x^3-14x-(4x^2y+8y)+6x^2y-12x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2x^3-14x-4x^2y-8y+6x^2y-12x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2x^3+2x^2y-12x^2-14x-8y\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2x} $ by $ \left( x^2-7\right) $ $$ \color{blue}{2x} \cdot \left( x^2-7\right) = 2x^3-14x $$Multiply $ \color{blue}{4y} $ by $ \left( x^2+2\right) $ $$ \color{blue}{4y} \cdot \left( x^2+2\right) = 4x^2y+8y $$$$ \left( \color{blue}{2y-4}\right) \cdot 3x^2 = 6x^2y-12x^2 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4x^2y+8y \right) = -4x^2y-8y $$ |
| ③ | Combine like terms: $$ 2x^3-14x \color{blue}{-4x^2y} -8y+ \color{blue}{6x^2y} -12x^2 = 2x^3+ \color{blue}{2x^2y} -12x^2-14x-8y $$ |
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