Using the rational root theorem, what are the possible rational roots of $x^{3} - 34 x + 12 = 0$ $x^3-34x+12=0$ ?

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Using the rational root theorem, what are the possible rational roots of $x^{3} - 34 x + 12 = 0$ $x^3-34x+12=0$ ?

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Answer 1 · 2016-10-06T06:25:51

According to the theorem, the possible rational roots are:

$\pm 1$ $+-1$ , $\pm 2$ $+-2$ , $\pm 3$ $+-3$ , $\pm 4$ $+-4$ , $\pm 6$ $+-6$ , $\pm 12$ $+-12$

Explanation:

$f (x) = x^{3} - 34 x + 12$ $f(x) = x^3-34x+12$

By the rational root theorem, any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integeres $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $12$ $12$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$ $+-1$ , $\pm 2$ $+-2$ , $\pm 3$ $+-3$ , $\pm 4$ $+-4$ , $\pm 6$ $+-6$ , $\pm 12$ $+-12$

Trying each in turn, we eventually find that:

$f (- 6) = {(- 6)}^{3} - 34 (- 6) + 12$ $f(color(blue)(-6)) = (color(blue)(-6))^3-34(color(blue)(-6))+12$

$f (- 6) = - 216 + 204 + 12$ $color(white)(f(color(white)(-6))) = -216+204+12$

$f (- 6) = 0$ $color(white)(f(color(white)(-6))) = 0$

So $x = - 6$ $x=-6$ is a rational root.

The other two roots are Real but irrational.

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Using the rational root theorem, what are the possible rational roots of $x^{3} - 34 x + 12 = 0$ $x^3-34x+12=0$ ?

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Using the rational root theorem, what are the possible rational roots of x3−34x+12=0x^3-34x+12=0 ?

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Using the rational root theorem, what are the possible rational roots of $x^{3} - 34 x + 12 = 0$ $x^3-34x+12=0$ ?