Answer :
step 1: According to the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n complex roots (counting multiplicities).
step 2:The given function f(x) is a polynomial of degree 5, so it can have up to 5 complex zeros.
step 3:To determine the type of zeros, we can look at the leading coefficient and the signs of the coefficients of the terms.
step 4:The leading coefficient is 1 (coefficient of x^5), which means the leading term is positive. This implies that there are no negative real zeros.
step 5:By Descartes’ Rule of Signs, we can determine the possible number of positive real zeros by counting the sign changes in the coefficients of f(x).
step 6:In f(x)=x5+9x3-2x^2+6x+5, there is only 1 sign change, indicating that there is exactly 1 positive real zero.
step 7:Since the total number of complex zeros is 5 and we have 1 positive real zero, the remaining zeros must be complex.
Final Answer:The function f(x)=x5+9x3-2x^2+6x+5 could have 1 positive real zero and 4 complex zeros.
step 2:The given function f(x) is a polynomial of degree 5, so it can have up to 5 complex zeros.
step 3:To determine the type of zeros, we can look at the leading coefficient and the signs of the coefficients of the terms.
step 4:The leading coefficient is 1 (coefficient of x^5), which means the leading term is positive. This implies that there are no negative real zeros.
step 5:By Descartes’ Rule of Signs, we can determine the possible number of positive real zeros by counting the sign changes in the coefficients of f(x).
step 6:In f(x)=x5+9x3-2x^2+6x+5, there is only 1 sign change, indicating that there is exactly 1 positive real zero.
step 7:Since the total number of complex zeros is 5 and we have 1 positive real zero, the remaining zeros must be complex.
Final Answer:The function f(x)=x5+9x3-2x^2+6x+5 could have 1 positive real zero and 4 complex zeros.
