Quadratic Equation

7x^3+x^2-7x-1=(7x+1)(x+1)(x-1)=0
x={-1/7,-1,1}

The equation is 7x^3+x^2-7x-1=0

Using the rational root theorum
all possible rational roots will be +-factors of -1 divided by +-factors of 7
that is +1, -1, +1/7, -1/7
They are the only posible RATIONAL roots
Sub x=+1 into the LHS

LHS=7+1-7-1 = 0 therefore x=1 is a solution so (x-1) is a factor
Try x=-1
LHS=-7+1+7-1=0 therefore x=-1 is alsot a solution so (x+1) is also a factor
Try x=1/7
LHS = 1/7 ²+1/7 ²-1+1 not 0 so not helpful
Try x=-1/7
LHS = -1/7 ²+1/7 ²+1-1 = 0 so x=-1/7 is a solution so x=-1/7 is a solution so (x+1/7) is also a factor
I know now that I have all the factors because when I multiply them all together the highest power of x is 3 which is correct

so
7x^3+x^2-7x-1 = k(x+1/7)(x-1)(x+1) so k=7 (because that's what is the coefficient of x^3)
7x^3+x^2-7x-1 = 7(x+1/7)(x-1)(x+1)

The solutions will be x=-1/7,x=1 and x=-1

Normally there are more than 4 possible rational roots (and often some other roots that are irrational) so I would normally just find one of them and then use synthetic division before looking for the next one. This you tube clip is on synthetic division.
http://www.youtube.com/watch?v=_zl63HL7Npw