Use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros for the polynomial
function.
-$f ( x ) = x ^ { 5 } + x ^ { 4 } + x ^ { 2 } + x + 7$
A)0 positive zeros, 1 negative zero
B)0 positive zeros, 0 negative zeros
C)0 positive zeros, 2 or 0 negative zeros
D)0 positive zeros, 3 or 1 negative zeros

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs for positive zeros counts sign changes in $f(x)$: $x^5 + x^4 + x^2 + x + 7$ has 0 sign changes, indicating 0 positive zeros. For negative zeros, substitute $x$ with $-x$ to get $f(-x) = (-x)^5 + (-x)^4 + (-x)^2 + (-x) + 7 = -x^5 + x^4 + x^2 - x + 7$, which has 3 sign changes, indicating 3 or 1 negative zeros (subtracting 2 each time until it's non-positive).

Use Descartes' Rule of Signs to Determine the Possible Number f(x)=x5+x4+x2+x+7f ( x ) = x ^ { 5 } + x ^ { 4 } + x ^ { 2 } + x + 7f(x)=x5+x4+x2+x+7

Use Descartes' Rule of Signs to Determine the Possible Number $f ( x ) = x ^ { 5 } + x ^ { 4 } + x ^ { 2 } + x + 7$