SOLUTION: how do you find the complex zeros of f(x)=x^3-8x^2+25x-26?

Question 415954: how do you find the complex zeros of f(x)=x^3-8x^2+25x-26?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
f(x)=x^3-8x^2+25x-26
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The best way to find all the roots of functions higher than second degree is to start by displaying this function on a graphing calculator to see if there are any real roots. As you can see from the graph below,the given cubic function has only one real root=2.
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Next, divide the function by (x+2),by long division or use synthetic division:
When you do this, you will get a quotient=x^2-6x+13
Solving this second degree equation using the quadratic formula:
a=1, b=-6, c=13
x=-(-6)+-sqrt((-6)^2-4*1*13)/2*1
=(6+-sqrt(36-52)/2
=(6+-sqrt(-16))/2
=(6+-4i)/2
=3+-2i
Ans:
one real root, 2
two complex roots,3+2i and 3-2i
See the graph below:
..

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