I have been struggling a little bit over solving cubic functions. I have been trying to use the P and Q method.
So the question is
What is the approximate value of the greatest zero of $f(x) = x^3 - 6x^2 - x + 3$
So I got the $P's$: $\{1,-1,3,-3\}$ and the $Q's$: $1,-1$. Now I used the $\frac pq $and got this values: $\{1,-1,3,-3\}$ and started to use synthetic division to get a remainder of $0$. However, none of those values gave me a remainder of $0$. What am I doing wrong?
Using Descartes rule of signs, there can be maximum two positive roots, and we observe the sign changes $f(0) = 3, f(6) = -3, f(7)=45$, so the largest root $\alpha \in (6, 7)$.
Writing $\alpha = 6+\epsilon$, we have $0< \epsilon < 1$, and we get $f(6+\epsilon) = -3+35\epsilon + 12\epsilon^2+\epsilon^3 = 0$. Ignoring the cubic term, we get $\epsilon = \frac1{12}$, so $\alpha \approx 6\frac1{12}$ should be a good approximation.