Use Descartes' rule of signs to determine the total number of real zeros and the number of positive and
negative real zeros. (Hint: First factor out x to its lowest power.)
-$$f ( x ) = 5 x ^ { 11 } + 2 x ^ { 9 } + 9 x ^ { 7 } - 9 x ^ { 6 }$$
A) 11 real zeros; f (x) has 1 positive real zero, 4 negative real zeros and the number 0 is a zero of multiplicity 6.
B) 6 real zeros; f (x) has 1 positive real zero, no negative real zeros and the number 0 is a zero of multiplicity 5.
C) 7 real zeros; f (x) has 1 positive real zero, no negative real zeros and the number 0 is a zero of multiplicity 6.
D) 2 real zeros; f (x) has 1 positive real zero, no negative real zeros and the number 0 is a zero of multiplicity 1.

Question

Quizplus · Accepted Answer

Descartes' rule of signs helps determine the number of positive and negative real zeros of a polynomial. For $f(x) = 5x^{11} + 2x^9 + 9x^7 - 9x^6$, factoring out $x^6$ gives $x^6(5x^5 + 2x^3 + 9x - 9)$. This indicates 6 zeros at $x = 0$. For positive real zeros, in $5x^5 + 2x^3 + 9x - 9$, there's 1 sign change, indicating 1 positive real zero. For negative real zeros, substitute $x = -x$ into the factored form and observe sign changes. There are no sign changes, indicating no negative real zeros. Thus, there are 7 real zeros in total: 1 positive, 0 negative, and 6 at zero.

Use Descartes' Rule of Signs to Determine the Total Number f(x)=5x11+2x9+9x7−9x6f ( x ) = 5 x ^ { 11 } + 2 x ^ { 9 } + 9 x ^ { 7 } - 9 x ^ { 6 }f(x)=5x11+2x9+9x7−9x6

Use Descartes' Rule of Signs to Determine the Total Number $f ( x ) = 5 x ^ { 11 } + 2 x ^ { 9 } + 9 x ^ { 7 } - 9 x ^ { 6 }$