Emily W. answered 01/09/22
High School and College Level Math and Science in Central Florida
f(x) = 2x^3 - 4x^2 + 8x - 3
First, check the number of times that the function f(x) changes sign. Here is the function broken apart:
+2x^3
-4x^2
+8x
-3
+ to - 1 change
- to + = 2 changes
+ to - = 3 changes
There are 3 sign changes in f(x). This is the maximum number of POSITIVE REAL solutions that can exist. We subtract 2 from this number until we pass 0. These are all the possible numbers of positive real solutions we can have
3 max + real solns
1 possible + real solns
-1 can’t have a negative amount of solutions
either 1 or 3 positive real solutions
Next we count the sign changes in f(-x) to figure out how many possible negative real solutions there can be. We are going to flip the sign of x ***NOT the coefficient** and see if it creates any changes to our function. Then we will recount the sign changes.
f(-x) = 2(-x)^3 - 4(-x)^2 + 8(-x) - 3
(-x)^3 = -x^3 so our first term changes to -2x^3
(-x)^2 is the same as x^2 so our 2nd term remains -4x^2
(-x) = -x so our third term changes to -8x
-3 remains the same
f(-x) = -2x^3 - 4x^2 -8x -3
All terms of f(-x) are negative so there is no sign change. There are no negative real solutions
0 negative real solutions
Finally, to determine the number of complex solutions, we consider each scenario for positive and negative real solutions, then find how many zeroes are unaccounted for. These will be the imaginary zeroes. Imaginary zeroes ALWAYS COME IN PAIRS, so you must always have either 0 or an even number of imaginary solutions.
Total number of zeroes = highest exponent = 3
Each row must add to 3
POSITIVE NEGATIVE IMAGINARY
3 0 0
1 0 2
We can have either 0 or 2 imaginary solutions