I am to use the rational zero theorem to find the real zeros of $2x^3-3x^2-x+1$.
The provided answers are $\frac{1}{2}$, $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$.
I was able to get $\frac{1}{2}$ on my own but I do not see how the other two were found.
In my textbook I am told I can find candidate 0's by taking the quotient of factors of p over factors of q where p is the constant term and q is the leading coefficient.
In this case p = factors of 1 = $\pm1$ and q = factors of 2 = $\pm1, 2$. Dividing combinations of p/q I get $\pm1$ and $\pm\frac{1}{2}$.
I then tried substituting for x each of those 4 combinations within the function $2x^3-3x^2-x+1$ and found that $\frac{1}{2}$ is a zero.
How/why are $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$ also zeros and how could I determine that from the rational zero theorem only? This last part is important since the chapter in my textbook and section of exercises specifically states I am to use this theorem to determine the real zeros.
Best Answer
Hint By rational root theorem you get that $1/2$ is a root to the cubic and so $2x-1$ is a factor of the cubic. Applying division algorithm, we get, $$2x^3-3x^2-x+1=(2x-1)(x^2-x-1)=0$$Now can you solve the quadratic $x^2-x-1=0$?