# Finding all real zeros of the polynomial?

Okay, so I need to find all real zeros in this polynomial...

$$f(x) = 2x^3 + x^2 - 13x + 6$$

I know that the first step is to find the factors of 6 and 2, then see which when multiplied by the other coefficients have them add up to equal zero, but none of the factors I tried came out to zero. Is there an easier way to go about doing this???

-
You might apply Cardano's formula. –  AD. Oct 5 '12 at 12:43
(Yes, that was a joke..) –  AD. Oct 5 '12 at 12:43

Your method will in general not find all real, but only all rational zeroes. If the leading coefficient were 1 instead of 2, all rational zeroes would have to be divisors of 6 (i.e. $\{\pm1, \pm2, \pm3, \pm6\}$). However with a leading coefficient of 2, one should also check halves of these values (i.e. also {$\pm\frac12, \pm\frac32\}$). Plugging in $x=2$, you will find that it is in fact a root. By polynomial division you thus obtain a quadratic for the other roots, which you can solve (or you will happen to find the remaining roots also by trying the above candidates).

-
Ugh, okay, for whatever reason I didn't consider halves as factors too. This makes far more sense now. Thanks! –  Brandt Oct 5 '12 at 13:46

If we assume we have only rational zeros we may the equation as $$2(x-a)(x-b)(x-c)=0= f(x)$$ with $a,b,c\in\mathbb{Q}$, expanding this leads to relations, which is a guide for guessing.

If there is only one rational solution we have $$(x-a)(2x^2-bx+c)=0= f(x)$$ with $a,b,c\in\mathbb{Q}$.

-
$$2x^{3}+x^{2}-13x+6=(x-2)(2x^{2}+5x-3)=(x-2)(x+3)(2x-1)$$ so the zeros are $-3$, $\frac{1}{2}$ and $2$.