Chapter 4.2, Problem 28AYU

In Problems 33-44, determine the maximum number of real zeros that each polynomial function may have. Then list the potential rational zeros of each polynomial function. Do not attempt to find the zeros.

40. $f\left(x\right)\text{ = -4}{x}^3\text{ + }{x}^2\text{ + }x\text{ + 6}$

To determine

Maximum Number of real positive or negative real zeros of the given polynomial function using Descartes’ Rule of signs and the potential rational zeros of each polynomial function using rational zeros theorem.

Answer to Problem 28AYU

The degree of the polynomial is 4. Therefore the number of real zeros by real zero theorem can be at most $=4$.

Therefore there will be 1 positive real zero.

Therefore there can be two negative real zeros or no negative real zero.

The rational zeros $=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6\pm \frac{1}{2},\pm \frac{1}{4}\pm \frac{3}{2}\pm \frac{3}{4}$.

Explanation of Solution

Given:

$f\left(x\right)=-4x^3+x^2+x+6$

The degree of the polynomial is 3. Therefore the number of real zeros by real zero theorem can be at most $=3$.

$f\left(x\right)=-4x^3+x^2+x+6$

$-$ to $+$

There is one variation of sign of nonzero coefficient of $f\left(x\right)$ by Descartes’ Rule of signs;

Number of positive real zeros $=$ number of variations of signs of $f\left(x\right)$ or number of variations less than even integer.

Therefore there will be 1 positive real zero.

Let us consider $f\left(-x\right)$ by replacing $x$ by $-x$.

$f\left(-x\right)=-4{(-x)}^3+{(-x)}^2+\left(-x\right)+6$

$f\left(-x\right)=4x^3+x^2-x+6$

$+$ to $-,-$ to $+$

There are two variations of signs of nonzero coefficient of $f\left(-x\right)$ by Descartes’ Rule of signs;

Number of negative real zeros $=$ number of variations of signs or number of variations less than even integer.

Therefore there can be two negative real zeros or no negative real zero.

Rational zeros theorem provides information about the rational zeros of a polynomial function with integer coefficients.

If $\frac{p}{q}$ in its lowest terms is a rational zero of $f$, then $p$ is a factor of $a_0$ and $q$ is the factor of $a_n$.

$f\left(x\right)=-4x^3+x^2+x+6$

Here $a_0=\text{ 6}$ and $a_n=-4$.

Zeros of $-4$, $q=\pm 1,\pm 2,\pm 4$.

Zeros of 9, $p=\pm 1,\pm 2,\pm 3,\pm 6$.

The potential rational zeros $=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6\pm \frac{1}{2},\pm \frac{1}{4}\pm \frac{3}{2}\pm \frac{3}{4}$.