Answer
$$(2k-5)(3k-4)=6k^2-23k+20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2k-5)(3k-4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6k^2-8k-15k+20 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6k^2-23k+20\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2k-5}\right) $ by each term in $ \left( 3k-4\right) $. $$ \left( \color{blue}{2k-5}\right) \cdot \left( 3k-4\right) = 6k^2-8k-15k+20 $$ |
| ② | Combine like terms: $$ 6k^2 \color{blue}{-8k} \color{blue}{-15k} +20 = 6k^2 \color{blue}{-23k} +20 $$ |
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