Since 1 is the constant and the only roots of 1 have a magnitude of 1, we can see that ax=1 or ax=-1 is a root, where a is a constant. If we put x=1 into the function, it does in fact equal zero. So x-1 is a factor, which we can divide by, using synthetic division: 6x^2+6x+1. Therefore f(x)=(x-1)(6x^2+6x+1). The quadratic implies that we have to look at the factors of 6 (that is 6 and 1, and 2 and 3) and the factors of 1 (that is 1 and 1, and -1 and -1) and combine these possibilities to get the coefficient of the middle term, which is 6. We also know that from the sign of the constant we have to add the products of the prospective factors, and we know from that that we can have no minus signs. So we're looking for 1 and 1 as the constants, and the products are 2*1+3*1=5 or 6*1+1*1=7. Neither of these gives us the 5 we need. The solution of the quadratic is x=(-6+sqrt(36-24))/12=(-3+sqrt(3))/6=-0.2113 or -0.7887. Therefore the three roots are 1, -0.2113, -0.7887 approx.