How do you solve #6x^3+x^2-6x-1=0#?

1 Answer
Massimiliano
Feb 5, 2015

The answer is: #x=+-1andx=-1/6#.

We can recollection as a partial common factor between the first and the third, and we can notice that the second and the fourth are a difference of squares.

So:

#6x(x^2-1)+(x^2-1)=0rArr(x^2-1)(6x+1)=0rArr#

#(x+1)(x-1)(6x+1)=0#, and than:

#x=+-1andx=-1/6#.

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