How do you use the rational roots theorem to find all possible zeros of $x^{3} - 3 x - 2$ $x^3-3x-2$ ?

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How do you use the rational roots theorem to find all possible zeros of $x^{3} - 3 x - 2$ $x^3-3x-2$ ?

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Answer 1 · 2016-06-02T18:03:26

$x^{3} - 3 x - 2$ $x^3-3x-2$ has zeros $x = - 1$ $x=-1$ with multiplicity $2$ $2$ and $x = 2$ $x=2$ .

Explanation:

$f (x) = x^{3} - 3 x - 2$ $f(x) = x^3-3x-2$

By the rational root theorem, any rational zeros of $f (x)$ $f(x)$ must be expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $- 2$ $-2$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$ $+-1$ , $\pm 2$ $+-2$

We find:

$f (- 1) = - 1 + 3 - 2 = 0$ $f(-1) = -1+3-2 = 0$

So $x = - 1$ $x=-1$ is a zero and $(x + 1)$ $(x+1)$ a factor of $f (x)$ $f(x)$ :

$x^{3} - 3 x - 2 = (x + 1) (x^{2} - x - 2)$ $x^3-3x-2 = (x+1)(x^2-x-2)$

$- 1$ $-1$ is also a zero of $x^{2} - x - 2$ $x^2-x-2$ :

If $x = - 1$ $x=-1$ then $x^{2} - x - 2 = 1 + 1 - 2 = 0$ $x^2-x-2 = 1+1-2 = 0$

Then we find:

$x^{2} - x - 2 = (x + 1) (x - 2)$ $x^2-x-2 = (x+1)(x-2)$

So $x = 2$ $x=2$ is the last zero.

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How do you use the rational roots theorem to find all possible zeros of $x^{3} - 3 x - 2$ $x^3-3x-2$ ?

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