Take the first coeff (6) and the last one (7) and form possible rational roots:
7/6, 7/3, 7, 2, 7/1. Check whether any of these constitutes a root of the polynomial. I will use synthetic division for this purpose. Are you familiar with synth. div.?
I tried using 7/1 (or just 7) as the divisor in synth div. There was no remainder. The coefficients of the quotient were 6 1 -1, representing
6x^2 + x -1. Looked like a possible rational root of 6x^2 + x - 1 is -1/6, but using -1/6 as the divisor in synth. div. resulted in a remainder of -1, meaning that -1/6 is not a root of the original equation.
I then used the quadratic formula to find the roots of 6x^2 + x - 1. They turned out to be 1/3 and - 1/2.
Thus, the zeros are {1/3, -1/2, 7}.
the factors are (x-1/3), (x+1/2), and (x-7).
Let me know if you want to review synthetic and/or long division.
