Question

According to the Rational Root Theorem, what are all the potential rational roots of f(x) = 9x^4 – 2x^2 – 3x + 4?

Answer 1

Answer:$\frac{p}{q}$ : ±1, ± $\frac{1}{3}$,  ± $\frac{1}{9}$, ±2 ±$\frac{2}{3}$,  ± $\frac{2}{9}$, ±4,  ± $\frac{4}{3}$

± $\frac{4}{9}$.

Step-by-step explanation:

Given: f(x) = $9x^4-2x^2-3x+4$ .

To find: According to the Rational Root Theorem, what are all the potential rational roots .

Solution: The rational theorem states that if f(x) has integer coefficients and $\frac{p}{q}$ is a rational zero, then q is the factor of leading coefficient and p is the factor of constant term.

                        f(x) = $9x^4-2x^2-3x+4$ .

Factor of leading coefficient (q) = ±9, ±3, ±1.

Factors of constant term( p)  =  ±4, ±2, ±1.

According to rational root theorem the rational roots are in the form $\frac{p}{q}$ : ±1, ± $\frac{1}{3}$,  ± $\frac{1}{9}$, ±2 ±$\frac{2}{3}$,  ± $\frac{2}{9}$, ±4,  ± $\frac{4}{3}$

± $\frac{4}{9}$.

Therefore , potential rational roots are $\frac{p}{q}$ : ±1, ± $\frac{1}{3}$,  ± $\frac{1}{9}$, ±2 ±$\frac{2}{3}$,  ± $\frac{2}{9}$, ±4,  ± $\frac{4}{3}$

± $\frac{4}{9}$.

Answer 2

According to the Rational Root Theorem, the potential rational roots of f(x) are$\pm 1,\pm \frac 13,\pm \frac 19,\pm 2, \pm \frac 23, \pm \frac 29, \pm 4, \pm \frac 43, \pm \frac 49$

The polynomial function is given as:

$f(x) = 9x^4 - 2x^2 - 3x + 4$

For a polynomial function

$f(x) = px^n +........+q$,

The possible rational roots are calculated using:

$Roots = \pm \frac{Factors\ of\ q}{Factors\ of p}$

So, we have:

$Roots = \pm \frac{Factors\ of\ 4}{Factors\ of 9}$

List the factors of 4 and 9

$Roots = \pm \frac{1,2,4}{1, 3, 9}$

Split

$Roots = \pm 1,\pm \frac 13,\pm \frac 19,\pm 2, \pm \frac 23, \pm \frac 29, \pm 4, \pm \frac 43, \pm \frac 49$

Hence, all the possible rational roots are: $\pm 1,\pm \frac 13,\pm \frac 19,\pm 2, \pm \frac 23, \pm \frac 29, \pm 4, \pm \frac 43, \pm \frac 49$

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