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Whoops i made an mistake on the previous one (2r+−5)(r+10) (2r)(r)+(2r)(10)+(−5)(r)+(−5)(10) \( \color{brown}{2r^2+20r−5r−50} \) \( \color{eneter color name}{2r^2+15r−50} \)
so 2r^2 +15r-50 is the answer
Yes, this is what i got
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