determine the max number of real zeros that p(x)=-6x^6+7x^5-4x^4+3x^2+9x-47 has
Real zeros can be found by using P over Q.
Factors of P over factors of Q.
You are very fortunate to have 47 as your P as its factors are only 1 and 47. As for Q, which is 6, its factors are 1,2,3,and 6. Do remember to put plus minus in front of each P divided by Q.
Use synthetic division for a problem such as this!
for synthetic division how do i know what to divide p(x)=-6x^6+7x^5-4x^4+3x^2+9x-47 by?
you're going to divide that long equation by the quotients you got from p/q.
you will have to keep trying until you get a remainder equal to 0. Also, when setting up the synthetic division, make sure you write +0x^3 between 4x^4 and 3x^2.
To determine the maximum number of real zeros that the polynomial function p(x) = -6x^6 + 7x^5 - 4x^4 + 3x^2 + 9x - 47 has, we can use the Descartes' Rule of Signs.
1. First, count the number of sign changes in the coefficients of the polynomial. In this case, we have:
-6x^6 + 7x^5 - 4x^4 + 3x^2 + 9x - 47
There are three sign changes: from -6x^6 to +7x^5, from +4x^4 to +3x^2, and from +3x^2 to -47x.
2. Next, find the number of positive real zeros by substituting (-x) for x and counting the sign changes in the coefficients. In this case, we have:
6x^6 + 7x^5 + 4x^4 + 3x^2 - 9x - 47
There are two sign changes: from 6x^6 to +7x^5, and from +4x^4 to -47x.
3. Now, subtract an even multiple of 2 from the number obtained in step 2. In this case, 2 is an even multiple of 2. So, we have 2 - 2 = 0.
Based on Descartes' Rule of Signs, there are either 0 or 2 positive real zeros.
4. Lastly, repeat steps 2 and 3 for f(-x) to find the number of negative real zeros. In this case, we have:
-6x^6 - 7x^5 - 4x^4 - 3x^2 + 9x - 47
There are three sign changes: from -6x^6 to -7x^5, from -4x^4 to -3x^2, and from -3x^2 to +9x.
Subtracting an odd multiple of 2 from the number obtained in step 4, we have 3 - (2*1) = 1.
According to Descartes' Rule of Signs, there is either 1 or 3 negative real zeros.
In conclusion, based on Descartes' Rule of Signs, the maximum number of real zeros that p(x) = -6x^6 + 7x^5 - 4x^4 + 3x^2 + 9x - 47 can have is either 0 or 2 positive real zeros and either 1 or 3 negative real zeros.