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 maximum number of zeros a polynomial can have is equal to its degree, which is 6 for this polynomial. Using Descartes' Rule of Signs, for positive zeros, we change $f(x)$ to $f(-x)$ and count sign changes. For $f(x)$, there are 2 sign changes, indicating 2 or 0 positive zeros. For $f(-x)$, there are 4 sign changes, 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 $f ( x ) = x ^ { 6 } - x ^ { 5 } - x ^ { 4 } - 5 x ^ { 3 } + 2 x ^ { 2 } + 2 x + 3$