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.
-$$f ( x ) = x ^ { 6 } - x ^ { 5 } - x ^ { 4 } - 5 x ^ { 3 } + 2 x ^ { 2 } + 2 x + 3$$
A) $6 ; 2$ or 0 positive zeros; 2 , or 0 negative zeros
B) 6; 3 or 1 positive zeros; 3 or 1 negative zeros
C) 6; 2 or 0 positive zeros; 4,2 , or 0 negative zeros
D) $6 ; 4,2$, or 0 positive zeros; 2 or 0 negative zeros

Question

Quizplus · Accepted Answer

The polynomial $f(x) = x^6 - x^5 - x^4 - 5x^3 + 2x^2 + 2x + 3$ has a maximum of 6 zeros because it is a 6th degree polynomial. Using Descartes' Rule of Signs, for positive zeros, we count the sign changes in $f(x)$: there are 2 sign changes ($-x^5$ to $-x^4$ and $2x^2$ to $2x$), indicating 2 or 0 positive zeros. For negative zeros, we substitute $x$ with $-x$ to get $f(-x) = (-x)^6 + (-x)^5 - (-x)^4 + 5(-x)^3 + 2(-x)^2 - 2x + 3$, which simplifies to $x^6 - x^5 - x^4 + 5x^3 + 2x^2 - 2x + 3$. There are 4 sign changes ($-x^5$ to $-x^4$, $-x^4$ to $5x^3$, $5x^3$ to $2x^2$, and $2x^2$ to $-2x$), indicating 4, 2, or 0 negative zeros.

Give the Maximum Number of Zeros the Polynomial Function May f(x)=x6−x5−x4−5x3+2x2+2x+3f ( x ) = x ^ { 6 } - x ^ { 5 } - x ^ { 4 } - 5 x ^ { 3 } + 2 x ^ { 2 } + 2 x + 3f(x)=x6−x5−x4−5x3+2x2+2x+3

Give the Maximum Number of Zeros the Polynomial Function May $f ( x ) = x ^ { 6 } - x ^ { 5 } - x ^ { 4 } - 5 x ^ { 3 } + 2 x ^ { 2 } + 2 x + 3$