Question
Answer
$$(2x+1)(3x-5)\cdot(4-x)=-6x^3+31x^2-23x-20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x+1)(3x-5)\cdot(4-x)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(6x^2-10x+3x-5)\cdot(4-x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(6x^2-7x-5)\cdot(4-x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}24x^2-6x^3-28x+7x^2-20+5x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-6x^3+31x^2-23x-20\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2x+1}\right) $ by each term in $ \left( 3x-5\right) $. $$ \left( \color{blue}{2x+1}\right) \cdot \left( 3x-5\right) = 6x^2-10x+3x-5 $$ |
| ② | Combine like terms: $$ 6x^2 \color{blue}{-10x} + \color{blue}{3x} -5 = 6x^2 \color{blue}{-7x} -5 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{6x^2-7x-5}\right) $ by each term in $ \left( 4-x\right) $. $$ \left( \color{blue}{6x^2-7x-5}\right) \cdot \left( 4-x\right) = 24x^2-6x^3-28x+7x^2-20+5x $$ |
| ④ | Combine like terms: $$ \color{blue}{24x^2} -6x^3 \color{red}{-28x} + \color{blue}{7x^2} -20+ \color{red}{5x} = -6x^3+ \color{blue}{31x^2} \color{red}{-23x} -20 $$ |
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