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Solve x^2 +3x-4=0
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\[0 = (x +4)(x-1)\]\[x=-4 and x= 1\]
(x+4)(x-1)=0 x = -4,1
\[x^{2}+3x-4=0\] Factor it. \[(x-1)(x+4)=0\] \[x=1, x=-4\]
x^2+3x-4 since a=1 b=3 c=-4 x^2+(a+b)x+(ab)=(x+a)(x+b) so you pick A and B which make those coefficients true -1+4=3 check (-1)(4)=-4 check so you can choose your A and B A=-1 B=4 so therefore (x-1)(x+4)=0 so x=1 and x=-4
disreguard the a=1 b=3 c=-4 does not have any sig in this explaination sorry
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