Chapter 2.5, Problem 23E

To determine

To find : the rational zeros of the given function

Answer to Problem 23E

The rational zeros of the function are: $\boxed{\boxed{-1,\&-6}}$

Explanation of Solution

Given information : $h\left(t\right)=t^3+8t^2+13t+6$

Concept Involved:

The Rational Zero Test : The Rational Zero Test relates the possible rational zeros of a polynomial (havinginteger coefficients) to the leading coefficient and to the constant term of the polynomial.

If the polynomial $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_2{x}^2+a_{1}x+a_0$ has integer coefficients, then every rational zero of $f$ has the formRational zero = $p/q$ where p and q have no common factors other than 1, $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$ .

To use the Rational Zero Test, you should first list all rational numbers whosenumerators are factors of the constant term and whose denominators are factors of theleading coefficient.

Possible rational zeros: $\frac{Factorsof\text{constant}term}{Factorsofleadingcoefficient}$

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.

Calculation:

Identifying the constant and leading coefficient of the given function

  $h\left(t\right)=t^3+8t^2+13t+6$

Constant is 6

Leading coefficient is 1

Listing the factors of constant 6: $\pm 1,\pm 2,\pm 3,\pm 6$

Listing the factors of Leading Coefficient 1: $\pm 1$

The possible rational zeros of the function are:

  $\pm 1,\pm 2,\pm 3,\&\pm 6$

Possible zero	Synthetic division	Is it a zero of polynomial?
$t=-1$	$\begin{array}{l}\begin{array}{l}-1\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ -1\end{array}\begin{array}{c}13\\ -7\end{array}\begin{array}{c}6\\ -6\end{array}}\\ 176\overline{)0}\leftarrow Remainder\end{array}$	Yes
$t=1$	$\begin{array}{l}\begin{array}{l}1\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ 1\end{array}\begin{array}{c}13\\ 9\end{array}\begin{array}{c}6\\ 22\end{array}}\\ 1922\overline{)28}\leftarrow Remainder\end{array}$	No
$t=-2$	$\begin{array}{l}\begin{array}{l}-2\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ -2\end{array}\begin{array}{c}13\\ -12\end{array}\begin{array}{c}6\\ -2\end{array}}\\ 161\overline{)4}\leftarrow Remainder\end{array}$	No
$t=2$	$\begin{array}{l}\begin{array}{l}2\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ 2\end{array}\begin{array}{c}13\\ 20\end{array}\begin{array}{c}6\\ 66\end{array}}\\ 11033\overline{)72}\leftarrow Remainder\end{array}$	No
$t=-3$	$\begin{array}{l}\begin{array}{l}-3\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ -3\end{array}\begin{array}{c}13\\ -15\end{array}\begin{array}{c}6\\ 6\end{array}}\\ 15-2\overline{)12}\leftarrow Remainder\end{array}$	No

Possible zero	Synthetic division	Is it a zero of polynomial?
$t=3$	$\begin{array}{l}\begin{array}{l}3\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ 3\end{array}\begin{array}{c}13\\ 33\end{array}\begin{array}{c}6\\ 138\end{array}}\\ 11146\overline{)144}\leftarrow Remainder\end{array}$	No
$t=-6$	$\begin{array}{l}\begin{array}{l}-6\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ -6\end{array}\begin{array}{c}13\\ -12\end{array}\begin{array}{c}6\\ -6\end{array}}\\ 121\overline{)0}\leftarrow Remainder\end{array}$	Yes
$t=6$	$\begin{array}{l}\begin{array}{l}6\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}8\\ 6\end{array}\begin{array}{c}13\\ 84\end{array}\begin{array}{c}6\\ 582\end{array}}\\ 11497\overline{)588}\leftarrow Remainder\end{array}$	No

Conclusion:

The possible rational zeros of the function are: $\pm 1,\pm 2,\pm 3,\pm 6$

The actual rational zeros of the function are: $-1,-6$