Answer
Fill the blanks with:
$5,\quad 3,\quad 1,\quad 0$
Work Step by Step
1. The number of positive real zeros of $P(x)$ is either equal to the number of variations in sign in $P(x)$ or is less than that by an even whole number.
2. The number of negative real zeros of $P(x)$ is either equal to the number of variations in sign in $P(-x)$ or is less than that by an even whole number.
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$P(x)=x^{5}-3x^{4}+2x^{3}-x^{2}+8x-8$
$P(-x)=-x^{5}-3x^{4}-2x^{3}-x^{2}-8x-8$
$P(x)$ has 5 sign changes so the number of positive zeros can be 5, 3, or 1.
$P(-x)$ has 0 sign changes so the number of negative zeros is 0.
