Question
Answer
$$(x+y-4)(x-4y-2)=x^2-3xy-4y^2-6x+14y+8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+y-4)(x-4y-2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2-3xy-4y^2-6x+14y+8\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x+y-4}\right) $ by each term in $ \left( x-4y-2\right) $. $$ \left( \color{blue}{x+y-4}\right) \cdot \left( x-4y-2\right) = x^2-4xy-2x+xy-4y^2-2y-4x+16y+8 $$ |
| ② | Combine like terms: $$ x^2 \color{blue}{-4xy} \color{red}{-2x} + \color{blue}{xy} -4y^2 \color{green}{-2y} \color{red}{-4x} + \color{green}{16y} +8 = x^2 \color{blue}{-3xy} -4y^2 \color{red}{-6x} + \color{green}{14y} +8 $$ |
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