Could you help me with this?
find the real zeros of
f(x) = x^3-8^2+9x+18(x-8^2+9x+18
I know it must be real simple. I have a block somewhere.
Thanks
I think I understand the Descartes rule of sign change.
This equation has 2 sign changes.
The total amount of real positive zeros could be 3 or 0.
I'm confused with negative real zeros and imaginary zeros
Thanks again
Dee
Dee
Of course, Dee! I'd be happy to help you with finding the real zeros of the given polynomial function and clarify your confusion with negative real zeros and imaginary zeros.
To find the real zeros of a polynomial function, we need to use a method called the Rational Root Theorem. This theorem states that if a rational number 𝑎/𝑏, where 𝑎 is a factor of the constant term and 𝑏 is a factor of the leading coefficient, is a zero (root) of the polynomial function, then 𝑎 is a factor of the constant term, and 𝑏 is a factor of the leading coefficient.
First, let's rewrite the given function in standard form:
f(x) = x^3 - 8x^2 + 9x + 18
Now, we can apply the Rational Root Theorem to find the possible rational zeros (roots) of the function. The constant term here is 18, and the leading coefficient is 1. Therefore, the possible rational zeros are all the factors of 18 divided by all the factors of 1.
The factors of 18 are:
±1, ±2, ±3, ±6, ±9, ±18
The factors of 1 are:
±1
Combining these factors, the possible rational zeros are:
±1, ±2, ±3, ±6, ±9, ±18
Next, we can use synthetic division or long division to test these possible zeros and see if any of them are indeed the zeros of the function. By doing so, we can narrow down the possibilities and find the real zeros.
Once we find any real zeros, we can use synthetic division to factorize the function and determine the remaining zeros (if any). Remember, if the function is factorable, it will have at least one linear factor. If not, then the remaining zeros could be either negative real zeros or imaginary zeros.
If you have any questions about the steps or need further assistance, feel free to ask!