Answer: [tex]\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}[/tex]
Step-by-step explanation:
We can use the Rational Root Test.
Given a polynomial in the form:
[tex]a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0[/tex]
Where:
- The coefficients are integers.
- [tex]a_n[/tex] is the leading coeffcient ([tex]a_n\neq 0[/tex])
- [tex]a_0[/tex] is the constant term [tex]a_0\neq 0[/tex]
Every rational root of the polynomial is in the form:
[tex]\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}[/tex]
For the case of the given polynomial:
[tex]2x^7+3x^5-9x^2+6=0[/tex]
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:
[tex]\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}[/tex]
