Respuesta :
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
[tex]f(x)=4x^4-2x^3-3x^2+6x-9[/tex]
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).
[tex]f(-x)=4(-x)^4-2(-x)^3-3(-x)^2+6(-x)-9[/tex]
[tex]f(-x)=4x^4+2x^3-3x^2-6x-9[/tex]
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.
