The function below has at least one rational zero.
Use this fact to find all zeros of the function.
h(x）=6x^3 - 31x^2 +4x+5
If there is more than one zero, separate them with commas. Write exact values, not decimal approximations
To find the rational zeros of the function h(x) = 6x^3 - 31x^2 + 4x + 5, we will use the Rational Root Theorem.
The Rational Root Theorem states that any rational zero of a polynomial function must be a factor of the constant term divided by a factor of the leading coefficient.
In this case, the constant term is 5 and the leading coefficient is 6. Therefore, the possible rational zeros are factors of 5 (1, 5) divided by factors of 6 (1, 2, 3, 6).
Possible rational zeros are: ±1, ±5, ±1/2, ±5/2, ±1/3, ±5/3, ±1/6, ±5/6
We can now use synthetic division or polynomial long division to test each possible rational zero.
Testing x = 1:
h(1) = 6(1)^3 - 31(1)^2 + 4(1) + 5 = 6 - 31 + 4 + 5 = -16
Since h(1) is not equal to 0, x = 1 is not a zero.
Testing x = -1:
h(-1) = 6(-1)^3 - 31(-1)^2 + 4(-1) + 5 = -6 - 31 - 4 + 5 = -36
Since h(-1) is not equal to 0, x = -1 is not a zero.
Testing x = 1/2:
h(1/2) = 6(1/2)^3 - 31(1/2)^2 + 4(1/2) + 5 = 3/2 - 31/4 + 2 + 5 = -3/4
Since h(1/2) is not equal to 0, x = 1/2 is not a zero.
Testing x = -1/2:
h(-1/2) = 6(-1/2)^3 - 31(-1/2)^2 + 4(-1/2) + 5 = -3/2 - 31/4 - 2 + 5 = -3/4
Since h(-1/2) is not equal to 0, x = -1/2 is not a zero.
Testing x = 1/3:
h(1/3) = 6(1/3)^3 - 31(1/3)^2 + 4(1/3) + 5 = 2/3 - 31/9 + 4/3 + 5 = 2/3 - 31/9 + 4/3 + 5 = 8/9
Since h(1/3) is not equal to 0, x = 1/3 is not a zero.
Testing x = -1/3:
h(-1/3) = 6(-1/3)^3 - 31(-1/3)^2 + 4(-1/3) + 5 = -2/3 - 31/9 - 4/3 + 5 = -8/9
Since h(-1/3) is not equal to 0, x = -1/3 is not a zero.
Testing x = 5/3:
h(5/3) = 6(5/3)^3 - 31(5/3)^2 + 4(5/3) + 5 = 50 - 775/9 + 20/3 + 5 = -20/9
Since h(5/3) is not equal to 0, x = 5/3 is not a zero.
Testing x = -5/3:
h(-5/3) = 6(-5/3)^3 - 31(-5/3)^2 + 4(-5/3) + 5 = -50 + 775/9 - 20/3 + 5 = 20/9
Since h(-5/3) is not equal to 0, x = -5/3 is not a zero.
The rational zeros of the function h(x) = 6x^3 - 31x^2 + 4x + 5 are not among the possible rational zeros we tested.
Therefore, h(x) = 6x^3 - 31x^2 + 4x + 5 does not have any rational zeros, and its zeros are irrational or complex.