Answer:
- complex zeros: -5+i, -5-i
- f(x) = (x -2)(x +5-i)(x +5+i)
Step-by-step explanation:
The first step is to divide the given cubic by the known factor to find the remaining quadratic factor. The attachment shows that being done by synthetic division. Then the complex roots will be the zeros of this quadratic.
f(x) = (x -2)(x^2 +10x +26)
We can rewrite the quadratic to vertex form* to simplify finding its zeros:
x^2 +10x +26 = (x^2 +10x +25) +1 = (x +5)^2 +1
Setting this to zero and solving for x, we get ...
(x +5)^2 +1 = 0
(x +5)^2 = -1
x +5 = ±√(-1) = ±i
x = -5 ±i . . . . . . . the complex zeros
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Of course, the linear factors are of the form (x-a) where "a" is a zero. You used that to find the quadratic factor. Factored to linear factors, we have ...
f(x) = (x -2)(x +5 -i)(x +5+i)
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* Alternatively, you can use the quadratic formula:
x = (-b±√(b²-4ac))/(2a) = (-10±√(10²-4·1·26))/(2·1)
= -5±(√-4)/2 = -5±i
