Lucille Davidson

2022-01-17

Determine the maximum number of real zeros that each polynomial function may have. Then use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros each polynomial function may have. Do not attempt to find the zeros.;
$f\left(x\right)-{x}^{3}-{x}^{2}+x+1$

Jack Maxson

Expert

Step 1: Given
$f\left(x\right)-{x}^{3}-{x}^{2}+x+1$
Step 2: Solution
Number of possible zeroes= degree
=3
$f\left(x\right)-{x}^{3}-{x}^{2}+x+1$
Number of times sign changes=2
So number of positive zeroes=2
$f\left(x\right)-{x}^{3}-{x}^{2}-x+1$
Number of times sign changes=1
Number of negative zeroes=1

soanooooo40

Expert

The maximum number of ral zeros of a polynomial function f is equal to the degree of f.
$f\left(x\right)=-{x}^{3}-{x}^{2}+x+1⇒$ The maximum number of real zeros is 3 since the degree is 3.
$f\left(x\right)=-{x}^{3}-{x}^{2}+x+1⇒$
Since there is only 1 change of signs in the coefficients of f(x), therefore the number of positive zeros must be 1.
$f\left(-x\right)={x}^{3}-{x}^{2}-x+1⇒$
Since there are 2 changes of signs in the coefficients of f(-x), therefore the number of negative zeros can be 2 or 0.

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