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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???

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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
up vote 6 down vote accepted

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).

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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 $a,b,c\in\mathbb{Q}$.

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find a number x that when you substitute in the polynomial will make it zero: the factor theorem. E.g when i substitute x = 2 , it gives zero. therefore (x-2) is a factor of the polynomial, then you perform polynomial division. when you divide 2x^2+x^2-13x+6 by (x-2) we get 2x^2+5x-3. then factorise 2x^2+5x-3 we get (x+3)(2x-1). therefore zeros are X=2,-3 and 1/2

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

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-1: While these are the zeroes, this doesn't help the OP learn how to solve problems. – Hurkyl Oct 5 '12 at 12:22

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