Give the maximum number of zeros the polynomial function may have. Use Descarte's Rule of Signs to determine how
many positive and how many negative zeros it may have.
-$$f ( x ) = x ^ { 7 } + x ^ { 6 } + x ^ { 2 } + x + 8$$
A) 7 ; 0 positive zeros; 1 negative zero
B) 7; 0 positive zeros; 2 or 0 negative zeros
C) $7 ; 0$ positive zeros; 3 or 1 negative zeros
D) 7; 0 positive zeros; 0 negative zeros

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs indicates the maximum number of positive and negative real zeros a polynomial can have. For positive zeros, we count the sign changes in $f(x)$. Here, there are no sign changes in $f(x) = x^7 + x^6 + x^2 + x + 8$, so there are 0 positive zeros. For negative zeros, we substitute $x$ with $-x$ and count the sign changes: $f(-x) = -x^7 - x^6 + x^2 - x + 8$, which has 3 sign changes, indicating there can be 3 or 1 negative zeros (subtracting 2 for each reduction in the maximum number of zeros).

Give the Maximum Number of Zeros the Polynomial Function May f(x)=x7+x6+x2+x+8f ( x ) = x ^ { 7 } + x ^ { 6 } + x ^ { 2 } + x + 8f(x)=x7+x6+x2+x+8

Give the Maximum Number of Zeros the Polynomial Function May $f ( x ) = x ^ { 7 } + x ^ { 6 } + x ^ { 2 } + x + 8$