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(will give medal) factor by grouping uy + 3u - 2y - 6
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(u-2)(y+3)
can you explain how you did it? (i'm having a major brain fart rn)
HI!!
it says "factor by grouping" right?
yes
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\[\color{red}{uy + 3u} -\color{blue}{ 2y - 6} \]
common factor of \(u\) in the red part, and \(2\) in the blue part
\[\color{red}{u(y+3)}+\color{blue}{-2(y+3)}\]
now each term has a common factor of \(y+3\) giving you \[(u-2)(y+3)\]
oh okay thank you that makes a lot more sense @misty1212
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