Use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros for the polynomial
function.
-$f ( x ) = - 4 x ^ { 5 } - 8 x ^ { 4 } - 4 x ^ { 3 } + 3 x ^ { 2 } + x + 16$
A)1 positive zero, 3 or 1 negative zeros
B)1 positive zero, 4 or 2 negative zeros
C)1 positive zero, 4 or 2 or 0 negative zeros
D)1 positive zero, 2 or 0 negative zeros

Question

Quizplus · Accepted Answer

For positive zeros, we count the sign changes in $f(x)$: there is only 1 sign change ($-4x^5$ to $+3x^2$), so there is 1 positive zero. For negative zeros, we substitute $x$ with $-x$ to get $f(-x) = -4(-x)^5 - 8(-x)^4 - 4(-x)^3 + 3(-x)^2 + (-x) + 16$, which simplifies to $4x^5 - 8x^4 + 4x^3 + 3x^2 - x + 16$. There are 3 sign changes here ($4x^5$ to $-8x^4$, $-8x^4$ to $4x^3$, and $3x^2$ to $-x$), so there could be 4, 2, or 0 negative zeros according to Descartes' Rule of Signs.

Use Descartes' Rule of Signs to Determine the Possible Number f(x)=−4x5−8x4−4x3+3x2+x+16f ( x ) = - 4 x ^ { 5 } - 8 x ^ { 4 } - 4 x ^ { 3 } + 3 x ^ { 2 } + x + 16f(x)=−4x5−8x4−4x3+3x2+x+16

Use Descartes' Rule of Signs to Determine the Possible Number $f ( x ) = - 4 x ^ { 5 } - 8 x ^ { 4 } - 4 x ^ { 3 } + 3 x ^ { 2 } + x + 16$