Question

What are the possible numbers of positive, negative, and complex zeros of f(x)=x^6+x^5+x^4+4x^3-12x^2+12?
Choices:
a. Positive: 2 or 0; Negative:4,2 or 0; Complex: 6, 4, 2 or 0
b. Positive:4, 2 or 0;Negative:2 or 0;Complex:4, 2 or 0
c. Positive:2 or 0;Negative:2 or 0;Complex:4, 2 or 0
d. Positive:2 or 0;Negative:0;Complex:6, 4, 2 or 0

I am thinking it's either a or c but I can't figure out which one.

Answer 1

I could really use help and I will give a medal if you tell me how. I am new!

Answer 2

You are counting the change in signs for f(x) and f(-x) between the terms.

So for f(x) you have this x^6 has a positive sign (+); x^5 has a positive sign (+); x^4 has a positive sign (+); 4x^3 has a positive sign; -12x has a negative sign (-); 12 has a positive sign (+)

So it's: ++++-+ There are two changes in sign, one between 4x^3 and -12x and one between -12x and 12.

This tells us that there either 2 positive roots and no complex roots or we have 0 positive roots and two complex roots. Note complex roots come in pairs.

Answer 3

Okay thank you! I actually already submitted what I needed to and it was a. I will give you a medal though for you since you helped me understand it more!

Answer 4

For f(-x) replace the x with -x.

So we have (-x)^6 + (-x)^5 + (-x)^4 + 4(-x)^3 - 12(-x)^2 +12

This becomes x^6 -x^5 +x^4 -4x^3 -12x +12 Notice that the terms with the odd exponent changed signs.

Again we go through the process of looking for changes in signs.

So x^6 has a positive sign (+); -x^5 has a negative sign (-) x^4 has a positive sign (+); -4x^3 has a negative sign (-); -12x has a negative sign; and 12 has a positive sign.

So we have +-+--+ or 4 changes.

So negative root will have either 4 negative real numbers or a multiple of two less then 4. So 4; 2 or 0 real negative roots. The complex roots can be either 4; 2 or 0

So the number of roots in a x^6 expression has to have a total of 6 roots.