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.
-$$f ( x ) = x ^ { 6 } + 4 x ^ { 5 } + 4 x ^ { 4 } + 3 x ^ { 3 } - x ^ { 2 } - 5 x + 4$$
A) 6 ; 2 or 0 positive zeros; 5,3 , or 1 negative zeros
B) 6; 3 or 1 positive zeros; 4 , 2, or 0 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 function given is of degree 6, so it can have a maximum of 6 zeros. Using Descartes' Rule of Signs for positive zeros, we count the number of sign changes in the original polynomial: $f(x) = x^6 + 4x^5 + 4x^4 + 3x^3 - x^2 - 5x + 4$. There are 2 sign changes, indicating 2 or 0 positive zeros. For negative zeros, we substitute $x$ with $-x$ and simplify to get $f(-x) = (-x)^6 + 4(-x)^5 + 4(-x)^4 + 3(-x)^3 - (-x)^2 - 5(-x) + 4 = x^6 - 4x^5 + 4x^4 - 3x^3 - x^2 + 5x + 4$, which has 4 sign changes, indicating 4, 2, or 0 negative zeros.

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

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