Please can someone explain the method to this with quadrinomials?
#Argand Diagrams
sterile horizon
How I’d do it
If 2+3i is a zero of f(z), then (z-(2+3i)) is a factor of f
If2+3i is a zero, then 2-3i is also a zero, and therefore z-(2-3i) is a factor of f
Find f(z) / [(z-(2+3i))(z-(2-3i))]
From there you get a quadratic which you can easy solve
drifting dagger
Better to do it the following way:
All the constants are real, so the roots are either real or complex conjugate. Thus, 3 - 2i is also a root, and thus we can find the other two roots...and 2 gee already wrote it.
sterile horizon
Oh wait sorry I misread it
drifting dagger
So yeah, do that.
sterile horizon
Not sure how you do it with variable coefficients tho
Also you probably could use the other zero we found to create a system of equations with a and b
dire osprey
so do f(0)?
dire osprey
right okay igy
drifting dagger
Not sure how you do it with variable coefficients tho
Why? It's obvious then.
First, we get (z - (2 + 3i))(z - (2 - 3i)) = z^2 - 4z + 13.
Dividing the 4th degree polynomial by a quadratic polynomial produces a second-degree polynomial. Besides, considering the leading coefficient of f(z) is 1 and the constant is 65, we get:
z^4 + az^3 + 6z^2 + bz + 65 = (z^2 - 4z + 13)(z^2 + cz + 5)
When we expand, we get:
z^4 + az^3 + 6z^2 + bz + 65 = z^4 + (c - 4)z^3 + (-4c + 18)z^2 + (13c - 20)z + 65
So, we get a system:
c - 4 = a
-4c + 18 = 6
13c - 20 = b
We only need the roots, so just find c from the second equation and solve z^2 + cz + 5 = 0, getting the remaining roots.
sterile horizon
Oh cool
drifting dagger
Though, your approach also works. But then the system will contain complex numbers - less friendly.
dire osprey
okay thanks both of you