Respuesta :

Rational root theorem:

Given a polynomial [tex]a_nx^n+a_{n-1}x^{n-1}+...+a_0[/tex], the rational roots are +-[tex] \frac{r_0}{r_n} [/tex] where [tex]r_0[/tex] = factor of the constant [tex]a_0[/tex] and [tex]r_n[/tex] = factors of the leading coefficient [tex]a_n[/tex].

So, to find the possible rational roots, we list all the factors of the constant and the leading coefficients and then set up the ratios.
In this case, the constant is [tex]a_0=–9[/tex] and the leading coefficient is [tex]a_n=1[/tex].
Factors of –9: +-1, +-3, +-9
Factors of 1: +-1

Thus, the possible rational roots are +-1/1, +-3/1, +-9/1
or +-1, +-3, +-9.

Answer: +-1, +-3, +-9

Using the Rational Roots(Zeros) Theorem, which states that, if the polynomial[tex]f(x) = a_nx^n + a_{n-1} x^{n-1} +...+ a_1x + a_0[/tex] has integer coefficients, then every rational zero of [tex]f(x)[/tex] has the form [tex]\dfrac{p}{q}[/tex] where [tex]p[/tex] is a factor of the constant term  [tex]a_0[/tex] and  [tex]q[/tex] is a factor of the leading coefficient  [tex]a_n.[/tex].

[tex]f(x) = x^3 + 2x - 9[/tex]

Here,

[tex]p: \pm1,\pm3, \pm9[/tex] which are all factors of constant term [tex]9[/tex]

[tex]q: \pm1[/tex] which are the factors of the leading coefficient [tex]1[/tex]

All possible values are

[tex]\dfrac{p}{q}: \pm1, \pm3, \pm9[/tex]

[tex]f(x) = x^3 + 2x - 9[/tex]

[tex]f(1) = 1 + 2 - 9 = -6[/tex]

[tex]f(-1) = -1 -2 -9 = -12[/tex]

[tex]f(3) = 27 + 6 -9 = 24[/tex]

No values are leading to zeros hence, none of these can be roots of the equation.

Learn more about rational root theorem here:

https://brainly.com/question/14232023

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