Question

determine the possible numbers of positive real zeros, and imaginary zeros for f(x)=2x^7-3x^6+5x^4+x^3-6X^2+1

Answer 1

Do you know Descartes's Rule of sign change? As the function stands, there are 4 changes of signs total (a change is when a positive sign changes to a negative sign or vice versa in consecutive signs. Like the first sign on the 2x^7 is positive, but the sign in front of the 3x^6 is a negative. That is one sign change. Here, there is a total of 4.)That means that there is a possibility of 4 positive real zeros out of the 7 possible (7 because the degree of the polynomial is 7. 2x^7). To find the number of negative real roots, you take f(-x), meaning you need to replace every x with a negative x and then see if the sign is still the same as it was. For example, in 2x^7, plug a -x in and see if its sign is positive or negative. -x^7 gives you a negative value. The next term has an x^6 in it. If you replace x with -x and take it to the 6th power, you have a positive sign, so a positive value of x times a negative value in front of the 3 still yields a -3x^6. Going on like this yields a possible 3 negative sign changes. The last and very important thing to know is that you can have a value of both positive and negative roots up to and including the highest number of sign changes and down by a difference of 2. I know that sound crazy but if we have 4 sign changes and get a high of 4 possible positive roots, we can also have 2 positive roots, or 0 positive roots. We find that we have 3 sign changes for the negative roots, but we can also have 1 because 3 - 2 = 1. The total we can have is 7. So if we have a combo of 4 positive and 3 negative, we can't have any imaginary cuz 4 + 3 = 7. Let's see how this looks in table form, cuz tables keep great track of stuff!