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

Answer: $\pm1,\frac{\pm1}{3},\frac{\pm1}{9},\pm2,\frac{\pm2}{3},\frac{\pm2}{9},\pm4,\frac{\pm4}{3},\frac{\pm4}{9}$

Step-by-step explanation:

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.

Given polynomial= $9x^4-2x^2-3x+4$

Now, Factors of leading coefficient q=$\pm9,\pm3,\pm1$

factors of constant term p=$\pm4,\pm2,\pm1$

according to rational root theorem the rational roots are in the form

$\frac{p}{q}=\pm1,\frac{\pm1}{3},\frac{\pm1}{9},\pm2,\frac{\pm2}{3},\frac{\pm2}{9},\pm4,\frac{\pm4}{3},\frac{\pm4}{9}$