Determine how many positive and how many negative real zeros the polynomial P(x)=x8−x5+x4−x3+x2−x+1 can have, using the Descartes' Rule of signs. Then determine the possible total number of real zeros.
We first look at P(x)
P(x)=+x8−x5+x4−x3+x2−x+1
P(x) has six variations in sign, so it has either 6,4,2 or positive roots.
Now,
P(−x)=(−x)8−(−x)5+(−x)4−(−x)3+(−x)2−(−x)+1P(−x)=+x8+x5+x4+x3+x2+x+1
So P(−x) has no variation in sign. Thus, P(x) has negative roots, making a total of either 7,5,3 or 1 real zeros. Since is a zero but is neither positive nor negative.
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