Expert: Richard J. Raridon - 7/6/2006

Question
1) How many imaginary zeros does the function    f(x)=(3x^4)+(2x^3)+(4x)+(7) have?

2)How many negative real zeros does the function f(x)=(x^5)+(x^4)-(x^2)-(x)+(1) have?

Can you please explain both? Thanks, I'm 14


Answer
1) you don't have to use parentheses.  You can write it
f(x) = 3x^4 +2x^3 +4x +7
If x=a is a solution to this equation then f(a) = 0.
You can easily see that no value of x > 0 is a solution.  
f(-1) = 4 and f(-2) = 31 and the values just keep getting bigger so no value of x < 0 is a solution.  Since imaginary values come in pairs, there must be 4 imaginary zeros.  
2) Again, you can write it f(x) = x^5 +x^4 -x^2 -x +1
There is no solution for x > 0
f(-1) = 1 and f(-2) = -17 so that tells you that there is a solution between -1 and -2.  f(-3) is even more negative so there are no more negative real zeros.  Therefore, you have 1 negative real zero and 4 imaginary zeros.
You're doing very well for age 14.