Chapter 2.5, Problem 25E

To determine

To find : the rational zeros of the given function

Answer to Problem 25E

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

Explanation of Solution

Given information:

  $C\left(x\right)=2x^3+3x^2-1$

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

  $C\left(x\right)=2x^3+3x^2-1$

Constant is -1

Leading coefficient is 2

Listing the factors of constant -1: $\pm 1$

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

The possible rational zeros of the function are:

  $\pm 1,\pm 1/2$

Possible zero	Synthetic division	Is it a zero of polynomial?
$x=-1$	$\begin{array}{l}\begin{array}{l}-1\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ -2\end{array}\begin{array}{c}0\\ -1\end{array}\begin{array}{c}-1\\ 1\end{array}}\\ 21-1\overline{)0}\leftarrow Remainder\end{array}$	Yes
$x=1$	$\begin{array}{l}\begin{array}{l}1\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ 2\end{array}\begin{array}{c}0\\ 5\end{array}\begin{array}{c}-1\\ 5\end{array}}\\ 255\overline{)4}\leftarrow Remainder\end{array}$	No
$x=-\frac{1}{2}$	$\begin{array}{l}\begin{array}{l}-1/2\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ -1\end{array}\begin{array}{c}0\\ -1\end{array}\begin{array}{c}-1\\ 1/2\end{array}}\\ 22-1\overline{)-1/2}\leftarrow Remainder\end{array}$	No
$x=\frac{1}{2}$	$\begin{array}{l}\begin{array}{l}1/2\\ \end{array}|\underset{\_}{\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ 1\end{array}\begin{array}{c}0\\ 2\end{array}\begin{array}{c}-1\\ 1\end{array}}\\ 242\overline{)0}\leftarrow Remainder\end{array}$	Yes

Conclusion:

The possible rational zeros of the function are:

  $\pm 1,\pm 1/2$

The actual rational zeros of the function are:

  $-1,1/2$