Answer :
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]
The possible rational roots of the polynomial equation [tex]0 = 2x^7 + 3x^5 - 9x^2 + 6.[/tex] are ±1, ±2, ±3, ±6, ±1/2, ±1, ±3/2, and ±3.
Given equation [tex]0 = 2x^7 + 3x^5 - 9x^2 + 6[/tex].
The Rational Root Theorem states that if there are any rational roots (fractions in the form p/q) for the polynomial equation, then p must be a factor of the constant term (in this case, 6), and q must be a factor of the leading coefficient (in this case, 2).
So, the possible rational roots (fractions p/q) for this polynomial are all the combinations of p and q, where:
p is a factor of 6, and
q is a factor of 2.
The factors of 6 are ±1, ±2, ±3, and ±6.
The factors of 2 are ±1 and ±2.
So, the possible rational roots are all the combinations of these factors:
±1/1, ±2/1, ±3/1, ±6/1, ±1/2, ±2/2, ±3/2, and ±6/2.
Simplify these fractions:
±1, ±2, ±3, ±6, ±1/2, ±1, ±3/2, and ±3.
Learn more about Rational Roots here:
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