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3r^2-16r-12=0 solve this quadratic equation. How do you factor this?
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use quadratic formula
x=\[(-b+-\sqrt{b ^{2}-4ac})/2a\]
wolfram says it is =(r-6)*(3r+2)=0 so r=6 or 3r=-2-->r=-2/3
i know the answer is 6, -2/3 but I don't understand how they came to that. The 3r^2 must be factor but it does not factor to 16.
use wht Hashir has written ...and tht's why i used Wolfram,cause i didn't know how to factorize it :/
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take all the factors of 3 and 12, and think of a combination that would give -16 which is your middle term -- Factors 3 ~ 3 and 1, factors of 12 ~ 3,4;2,6;12,1. only 3,1 and 2, -6 will fit that. Thats why, 3r^2 -16r -12 = (3r +2)(x -6) since when multiplying (3r +2)(x -6) will give you 3x^2 +2x -18x -12.
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