Question
Answer
$$(3x-1)(x-4)\cdot2=6x^2-26x+8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3x-1)(x-4)\cdot2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(3x^2-12x-x+4)\cdot2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(3x^2-13x+4)\cdot2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}6x^2-26x+8\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3x-1}\right) $ by each term in $ \left( x-4\right) $. $$ \left( \color{blue}{3x-1}\right) \cdot \left( x-4\right) = 3x^2-12x-x+4 $$ |
| ② | Combine like terms: $$ 3x^2 \color{blue}{-12x} \color{blue}{-x} +4 = 3x^2 \color{blue}{-13x} +4 $$ |
| ③ | $$ \left( \color{blue}{3x^2-13x+4}\right) \cdot 2 = 6x^2-26x+8 $$ |
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