melissa.ross23 said:
Determine the possible # of positive zeros and possible # of negative zeros. p(x)=3x^5-2x^4-5x^3+x^2-x-9
The ONLY tricky part is the understanding of Complex roots. Since we have Real coeficients, Complex roots must apear in pairs.
DeCartes' rule counts the MAXIMUM number of Real roots. If the actual number of Real roots is less than the maximum, it must be less by some factor of 2.
p(x)=3x^5-2x^4-5x^3+x^2-x-9
We have +--+--. This is three (3) sign changes. Start with whatever is first and count from there. We construe, then, that there MUST be One (1) positive Real root, but there may be three (3)
Likewise, p(-x)=-3x^5-2x^4+5x^3+x^2+x-9
This gives, --+++-. This is two (2) sign changes. It is sometimes disappointing to get an even number. We cannot, therefore, state that there MUST be a Negative Real root, only that there MAY be two (2).
That's really about it.
As it turns out, this p(x) has one Real root, the positive one we suggested above. The other four (4) are complex.
The Real root
1.81545
The Complex Roots
0.50239 + 0.95687i
0.50239 - 0.95687i
-1.07679 + 0.50528i
-1.07679 - 0.50528i
See how beautifully they appear in conjugate pairs?
