What are the possible number of positive, negative, and complex zeros of f(x) = 2x^3 – 5x^2 – 6x + 4 ? and use complete sentences to explain the method used to solve this equation.

Question

dumbcow · Answer

Its Descartes Rule of signs
count the change in signs for f(x), this will give you the number of positive zeroes
count the change in signs for f(-x) this will give the number of negative zeroes
The total must be the degree of the polynomial, so thats where complex zeroes come in and they must be in pairs

dumbcow · Answer

For example, look at sign of each coefficient
+,-,-,+
There are 2 instances where the sign changed, there could be 0 or 2 positive zeroes

f(-x) = -2x^3 -5x^2 +6x +4
-,-,+,+
There is 1 instance where the sign changed, there is 1 negative zero

0+1 = 1, but there must be total of 3 so there could be 2 complex zeroes as well
2+1 = 3, 0 complex zeroes in this situation

anonymous · Answer

don't you love these?  especially since it doesn't tell you what you actually have, just what you might have

anonymous · Answer

can you explain the complex numbers a little better, I don't really understand and thank you so much

dumbcow · Answer

complex zeroes come in pairs, you could have 0,2,4 ...
A complex zero is not a positive or negative zero
Since this is a cubic it must have a total of 3 zeroes
we also know it has exactly 1 negative zero
so the possible combinations are:
Positive       negative        complex
   0                1                    2
    2                1                   0

dumbcow · Answer

to know for sure, you would have to graph the function or factor it

anonymous · Answer

alright thank you so much :)