Chapter 2.5, Problem 98E

To determine

To find : the rational zeros of the given function

Answer to Problem 98E

The rational zeros of the function are: $\boxed{\boxed{z=-2,-\frac{1}{3},\&\frac{1}{2}}}$

Explanation of Solution

Given information: $f\left(z\right)=z^3+\frac{11}{6}{z}^2-\frac{1}{2}z-\frac{1}{3}$

Concept Involved:

The zeros of the polynomial are the values of x where the graph cuts the x-axis and makes $P\left(x\right)=0$

Synthetic Division (for a Cubic Polynomial):To divide $a{x}^3+b{x}^2+cx+d$ by $x-k$ , use this pattern.

$\begin{array}{l}\begin{array}{l}k\\ \end{array}|\stackrel{Coefficientofdividend}{\overbrace{\underset{\_}{\begin{array}{cccc}\boxed{a}& b& c& d\\ \downarrow & \underset{\downarrow }{\boxed{ka}}& \underset{\downarrow }{\boxed{}}& \underset{\downarrow }{\boxed{}}\end{array}}}}\\ \begin{array}{ccccc}& \nearrow & \nearrow & \nearrow & \end{array}\\ \underset{Coefficientofquotient}{\underbrace{\begin{array}{ccc}\boxed{a}& \boxed{}& \boxed{}\end{array}}}\begin{array}{cc}\boxed{r}& \leftarrow \text{Remainder}\end{array}\end{array}$

In case when we have a polynomial with a missing term, insert placeholders with zero coefficients for missing powers of the variable. Vertical pattern: Add terms in columns Diagonal pattern: Multiply results by k. This algorithm for synthetic division works only for divisors of the form x - k. Remember that $x+k=x-\left(-k\right)$ .

The Division Algorithm: If $f\left(x\right)$  and   $d\left(x\right)$ are polynomials such that $d\left(x\right)\ne 0$ , and the degree of $d\left(x\right)$ isless than or equal to the degree of $f\left(x\right)$ , then there exist unique polynomials $q\left(x\right)$  and   $r\left(x\right)$ such that $\underset{\stackrel{­}{{\displaystyle Dividend}}}{{\displaystyle f(x)}}=\underset{\stackrel{­}{{\displaystyle Divisor}}}{{\displaystyle d(x)}}\times \underset{\stackrel{­}{{\displaystyle Quotient}}}{{\displaystyle q(x)}}+\underset{\stackrel{­}{{\displaystyle Remainder}}}{{\displaystyle r(x)}}$ where $r\left(x\right)=0$ or the degree of $r\left(x\right)$ is less than the degree of $d\left(x\right)$ . If theremainder $r\left(x\right)$ is zero, then $d\left(x\right)$ divides evenly into $f\left(x\right)$ and k can be called as zero of the polynomial.

Graph:

  

Interpretation:

From the graph of the function we can pick possible zeros of the function as $z=-2,-\frac{1}{3},\&\frac{1}{2}$ and check using synthetic division whether they are actual zeros.

Possible zero	Synthetic division	Is it a zero of polynomial?
$z=-2$	$\begin{array}{l}\begin{array}{l}-2\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}11/6\\ -2\end{array}\begin{array}{c}-1/2\\ 1/3\end{array}\begin{array}{c}-1/3\\ 1/3\end{array}}\\ 1-1/6-1/6\overline{)0}\leftarrow Remainder\end{array}$	Yes
$z=-\frac{1}{3}$	$\begin{array}{l}\begin{array}{l}-1/3\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}11/6\\ -1/3\end{array}\begin{array}{c}-1/2\\ -1/2\end{array}\begin{array}{c}-1/3\\ 1/3\end{array}}\\ 13/2-1\overline{)0}\leftarrow Remainder\end{array}$	Yes
$z=\frac{1}{2}$	$\begin{array}{l}\begin{array}{l}1/2\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}11/6\\ 1/2\end{array}\begin{array}{c}-1/2\\ 7/6\end{array}\begin{array}{c}-1/3\\ 1/3\end{array}}\\ 17/32/3\overline{)0}\leftarrow Remainder\end{array}$	Yes

Conclusion:

The rational zeros of the function are: $z=-2,-\frac{1}{3},\&\frac{1}{2}$