Question

What are the possible rational roots of the polynomial equation? 0=2x7+3x5−9x2+6

Answer 1

Answer: $\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

Step-by-step explanation:

We can use the Rational Root Test.

Given a polynomial in the form:

$a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0$

Where:

- The coefficients are integers.

- $a_n$ is the leading coeffcient ($a_n\neq 0$)

- $a_0$ is the constant term $a_0\neq 0$

Every rational root of the polynomial is in the form:

$\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}$

For the case of the given polynomial:

$2x^7+3x^5-9x^2+6=0$

We can observe that:

- Its constant term is 6, with factors 1, 2 and 3.

- Its leading coefficient is 2, with factors 1 and 2.

 Then, by Rational Roots Test we get the possible rational roots of this polynomial:

$\frac{p}{q}=\frac{\pm(1,2,3,6)}{\pm(1,2)}=\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$