To find the rational roots (zeros) of the cubic equation 5x^3 + 6x^2 - 9x + 2 = 0, we can use the rational root theorem and synthetic division. The rational root theorem states that any rational root of the polynomial equation must be of the form p/q, where p is a factor of the constant term (2 in this case) and q is a factor of the leading coefficient (5 in this case).
Step 1: Find the factors of the constant term (2):
Factors of 2: ±1, ±2
Step 2: Find the factors of the leading coefficient (5):
Factors of 5: ±1, ±5
Step 3: Possible rational roots:
±1, ±2, ±1/5, ±2/5
Now, we'll test these potential roots using synthetic division:
1. For x = 1:
5(1)^3 + 6(1)^2 - 9(1) + 2 = 4 (not equal to 0)
2. For x = -1:
5(-1)^3 + 6(-1)^2 - 9(-1) + 2 = 10 (not equal to 0)
3. For x = 2:
5(2)^3 + 6(2)^2 - 9(2) + 2 = 30 (not equal to 0)
4. For x = -2:
5(-2)^3 + 6(-2)^2 - 9(-2) + 2 = -2 (not equal to 0)
5. For x = 1/5:
5(1/5)^3 + 6(1/5)^2 - 9(1/5) + 2 = 0 (equals 0)
Therefore, x = 1/5 is a rational root (zero) of the cubic equation 5x^3 + 6x^2 - 9x + 2 = 0. The other potential roots (±1, ±2, ±2/5) are not rational roots of this equation.