The RZT says that any rational root will be a factor of 22. So, the number of possibilities is quite limited:
±1 ±2 ±11 ±22
A little synthetic division shows that
f(x) = (x-11)(x+1)(x+2)
f(x)=x^3-8x^2-31x-22
±1 ±2 ±11 ±22
A little synthetic division shows that
f(x) = (x-11)(x+1)(x+2)
The factors of -22 are: -22, -11, -2, -1, 1, 2, 11, 22.
The factors of 1 are: 1.
So, the possible rational zeros are: ±1, ±2, ±11, ±22.
To determine if any of these values are actually zeros of f(x), we can substitute them into the polynomial function and see if the result is equal to 0.
Let's test each of the possible rational zeros:
For x = 1: f(1) = 1^3 - 8(1)^2 - 31(1) - 22 = -60 ≠0
For x = -1: f(-1) = (-1)^3 - 8(-1)^2 - 31(-1) - 22 = -4 ≠0
For x = 2: f(2) = 2^3 - 8(2)^2 - 31(2) - 22 = -28 ≠0
For x = -2: f(-2) = (-2)^3 - 8(-2)^2 - 31(-2) - 22 = -16 ≠0
For x = 11: f(11) = 11^3 - 8(11)^2 - 31(11) - 22 = -1200 ≠0
For x = -11: f(-11) = (-11)^3 - 8(-11)^2 - 31(-11) - 22 = 4 ≠0
For x = 22: f(22) = 22^3 - 8(22)^2 - 31(22) - 22 = -666 ≠0
For x = -22: f(-22) = (-22)^3 - 8(-22)^2 - 31(-22) - 22 = 20 ≠0
None of the possible rational zeros is an actual zero for f(x).
Since we haven't found any rational zeros, we need to use other methods to find the real zeros and factor the polynomial.
One way to find the real zeros is by using synthetic division with a guessed zero. By trying different values, we can eventually find a zero.
Alternatively, we can use numerical methods, such as graphing the function and finding the x-intercepts or using a calculator or software to solve the equation.
Let me know if you want to proceed with any of these methods, or if you have any other questions.