Chapter 2.5, Problem 28E

To determine

To find : the rational zeros of the given function

Answer to Problem 28E

The rational zeros of the function are: $\boxed{\boxed{1,-1,5,\&\frac{5}{2}}}$

Explanation of Solution

Given information:

  $f\left(x\right)=2x^4-15x^3+23x^2+15x-25$

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 $\underset{\stackrel{\uparrow }{{\displaystyle Dividend}}}{{\displaystyle \text{f (x)}}}\text{ = }\underset{\stackrel{\uparrow }{{\displaystyle Divisor}}}{{\displaystyle \text{d(x)}}}\cdot \underset{\stackrel{\uparrow }{{\displaystyle Quotient}}}{{\displaystyle \text{q(x)}}}+\underset{\stackrel{\uparrow }{{\displaystyle Remainder}}}{{\displaystyle \text{ r (x)}}}$ 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

  $f\left(x\right)=2x^4-15x^3+23x^2+15x-25$

Constant is -25

Leading coefficient is 2

Listing the factors of constant 24: $\pm 1,\pm 5,\pm 25$

Listing the factors of Leading Coefficient 1: $\pm 1,\pm 2$

The possible rational zeros of the function are: $\pm 1,\pm \frac{1}{2},\pm 5,\pm \frac{5}{2},\pm 25,\pm \frac{25}{2}$

Graph:

  

Interpretation:

From the graph of the function we can pick possible zeros of the function as

  $x=1,-1,5,\&\frac{5}{2}$ and check using synthetic division whether they are actual zeros.

Possible zero	Synthetic division	Is it a zero of polynomial?
$x=-1$	$\begin{array}{l}\begin{array}{l}-1\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}-15\\ -2\end{array}\begin{array}{c}23\\ 17\end{array}\begin{array}{c}15\\ -40\end{array}\begin{array}{c}-25\\ 25\end{array}}\\ 2-1740-25\overline{)0}\leftarrow Remainder\end{array}$	Yes
$x=1$	$\begin{array}{l}\begin{array}{l}1\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}-15\\ 2\end{array}\begin{array}{c}23\\ -13\end{array}\begin{array}{c}15\\ 10\end{array}\begin{array}{c}-25\\ 25\end{array}}\\ 2-131025\overline{)0}\leftarrow Remainder\end{array}$	Yes
$x=5$	$\begin{array}{l}\begin{array}{l}5\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}-15\\ 10\end{array}\begin{array}{c}23\\ -25\end{array}\begin{array}{c}15\\ -10\end{array}\begin{array}{c}-25\\ 25\end{array}}\\ 2-5-25\overline{)0}\leftarrow Remainder\end{array}$	Yes
$x=\frac{5}{2}$	$\begin{array}{l}\begin{array}{l}5/2\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}-15\\ 5\end{array}\begin{array}{c}23\\ -25\end{array}\begin{array}{c}15\\ -5\end{array}\begin{array}{c}-25\\ 25\end{array}}\\ 2-10-210\overline{)0}\leftarrow Remainder\end{array}$	Yes

Conclusion:

The possible rational zeros of the function are: $\pm 1,\pm \frac{1}{2},\pm 5,\pm \frac{5}{2},\pm 25,\pm \frac{25}{2}$

The actual rational zeros of the function are: $1,-1,5,\frac{5}{2}$