algebra 2 problem

Question

Could someone explain and or tell me what the real and negative zero are for this
f(x)= 3x^5-3x^2+x-5

Answer 1

by Descartes sign rule there are 1 or 3 + real zeros

and 0 - zeros, so 2 or 4 imaginary zeros. a 5th degree function has 5 zeros, and at least 1 real zero. all odd degree functions have at least 1 real zero

f(x) = 3x^5 -3x^2 +x -5

x f(x)

0 f(0)=-5

1 f(1) = 3-3+1-5 = -4

2 f(2) = 96- 12+2-5 = 81. one zero is between x=1 and 2, closer to 1, given the change in sign of f(x) from -4 to +81

x= about 1.225045 is the x intercept when you graph the function. It's the only x intercept = the only real zero. the other 4 zeros are imaginary. there is No negative real zero

x= about 1.23 rounded to 2 decimals