Answer:
So possibilities of zeroes are:
Positive Negative Imaginary
1 1 2
3 1 0
Zeroes = -1.4549, 1.2658, 0.34457-1.0503i, 0.34457+1.0503i.
Step-by-step explanation:
Note: Descartes' Rule of Signs is used to find the signs of zeroes not the exact value.
The given function is
Degree of polynomial is 4 so number of zeroes is 4.
There are three sign changes, so there are either 3 positive zeros or 1 positive zero.
Now, put x=-x in f(x).
There is one variation in sign change, so there is 1 negative zero.
So possibilities of zeroes are:
Positive Negative Imaginary
1 1 2
3 1 0
Using graphing calculator the zeroes of given function are -1.4549, 1.2658, 0.34457-1.0503i and 0.34457+1.0503i.