Question
Answer
$$2x(x-4)(3x+4)=6x^3-16x^2-32x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2x(x-4)(3x+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(2x^2-8x)(3x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6x^3+8x^2-24x^2-32x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}6x^3-16x^2-32x\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2x} $ by $ \left( x-4\right) $ $$ \color{blue}{2x} \cdot \left( x-4\right) = 2x^2-8x $$ |
| ② | Multiply each term of $ \left( \color{blue}{2x^2-8x}\right) $ by each term in $ \left( 3x+4\right) $. $$ \left( \color{blue}{2x^2-8x}\right) \cdot \left( 3x+4\right) = 6x^3+8x^2-24x^2-32x $$ |
| ③ | Combine like terms: $$ 6x^3+ \color{blue}{8x^2} \color{blue}{-24x^2} -32x = 6x^3 \color{blue}{-16x^2} -32x $$ |
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