SOLUTION: Can you help me with this problem? I can't remember of its completing the square or synthetic division or something else... I have a lot of others to do so could you walk me throug

Question 614897: Can you help me with this problem? I can't remember of its completing the square or synthetic division or something else... I have a lot of others to do so could you walk me through this one so I can try to figure out the rest? Thanks!
6x^4+7x^3-37x^2-8x+12=0 solve for x.
Thank you so much!!
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
6x^4+7x^3-37x^2-8x+12=0 solve for x.
using rational roots theorem to solve:
...0...|.....6.....7.....-37......-8......12
...1...|.....6....13....-24......-32..-20(There is a root between 0 and 1)(change of sign)
...2...|.....6....19......1........-6......0 (2 is a root)
...3...|.....6....25.....38.....106...330 (3 is upper boundary)(numbers all positive)
=====================
...0...|.....6....19......1.......-6
.-1...|.....6....13...-13........7(There is a root between 0 and -1)(change of sign)
.-2...|.....6....7.....-13......20
.-3...|.....6....1......-2.......0 (-3 is a root)
.-4...|.....6....-5......21.....-90 (4 is lower boundary)(numbers alternate in sign)
(x-2)(x+3)(6x^2+x-2)=0
(x-2)(x+3)(2x-1)(3x+2)=0
x=-3,-2/3,1/2,2
..
note: you can check your answers with a graphing calculator

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