Finding zeroes of $x^3-5x^2+11x+17$

Question

I'm trying to find all the zeros of $x^3-5x^2+11x+17$. I figured the possible zeros as being +/- 1, +/- 17$. The book says that -1 is supposed to be a factor, but I tried dividing the polynomial by -1 with synthetic division and it ended up not being a zero. I know I have to get it down to a degree of 2 to get the other answer, but from what I've seen I need to first find that other zero and use the result of the synth division which will put it at a degree of two, or something like that. SO I'm not really sure what to do with this now.

Answer 1

We can use the identities of Viète to avoid the hazards of polynomial division.

The sum of the roots is $5$, so the sum of the missing roots is $6$.

The product of the roots is $-17$, so the product of the missing roots is $17$.

It follows that the missing roots are solutions of $x^2-6x+17=0$.

Answer 2

Note: Let $f(x)=x^3-5x^2+11x+17$.

$f(-1)=0\Rightarrow (x+1)|f(x)$.

Divide, and get a quadratic factor. I'm sure you'll be able to find the other two roots then.