SOLUTION: Please help me figure out how to find the complex zeros of the equation: f(x)=4x^3+19x^2-61x+14

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Question 322743: Please help me figure out how to find the complex zeros of the equation:
f(x)=4x^3+19x^2-61x+14
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Since there are three roots and complex roots come in pairs, at least one must be real.
+graph%28+300%2C+300%2C+-10%2C+5%2C+-500%2C+500%2C+4x%5E3%2B19x%5E2-61x%2B14%29
All three roots are real for this equation.
Looks like x=-7 and x=2 are roots.
Verify using the equation.
I'll do one, you verify the other.
4x%5E3%2B19x%5E2-61x%2B14=4%28-7%29%5E3%2B19%2849%29-61%28-7%29%2B14
4x%5E3%2B19x%5E2-61x%2B14=-1372%2B931%2B427%2B14
4x%5E3%2B19x%5E2-61x%2B14=0
.
.
.
Now that you have two factors.
%28x%2B7%29%28x-2%29=x%5E2%2B5x-14
You can use polynomial long division to find the remaining factor.
%284x%5E3%2B19x%5E2-61x%2B14%29%2F%28x%5E2%2B5x-14%29