SOLUTION: Hello! I need help solving this: I have the function:f(x) = 4x^3 + 5x^2 + 36x + 45 and the zero: 3i I have to use the given zero to find all the zeros of the function. T

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Hello! I need help solving this: I have the function:f(x) = 4x^3 + 5x^2 + 36x + 45 and the zero: 3i I have to use the given zero to find all the zeros of the function. T      Log On


   

Question 977052: Hello!
I need help solving this:
I have the function:f(x) = 4x^3 + 5x^2 + 36x + 45
and the zero: 3i
I have to use the given zero to find all the zeros of the function.
Thank you
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Since it has rational coefficients, the complex conjugate of 3i is also a root.
So %28x-3i%29%28x%2B3i%29=x%5E2%2B9 is a factor of f(x).
Divide the factor from the polynomial to find the remaining factor.
%284x%5E3%2B5x%5E2%2B36x%2B45%29%2F%28x%5E2%2B9%29
You know that 4x is the leading term of the factor because of the 4x%5E3 leading term in f(x).
4x%28x%5E2%2B9%29=4x%5E2%2B36x
Which leaves,
4x%5E3%2B5x%5E2%2B36x%2B45-%284x%5E2%2B36x%29=5x%5E2%2B45
Then the constant term of the factor would be 5 since,
5%28x%5E2%2B9%29=5x%5E2%2B45
So the remaining factor is, 4x%2B5
%284x%5E3%2B5x%5E2%2B36x%2B45%29=%28x%5E2%2B9%29%284x%2B5%29
%284x%5E3%2B5x%5E2%2B36x%2B45%29=%28x%2B3i%29%28x-3i%29%284x%2B5%29
4x%2B5=0
4x=-5
x=-5%2F4
So the zeros are 3i,-3i, and -5/4.