Question 4.4.1: Using the Rational Zeros Theorem Find all rational zeros of...
Using the Rational Zeros Theorem
Find all rational zeros of F(x) = 2x³ + 5x² – 4x – 3.
First, we list all possible rational zeros of F(x).
\text { Possible rational zeros }=\frac{p}{q}=\frac{\text { Factors of the constant term, }-3}{\text { Factors of the leading coefficient, } 2}
Factors of -3: \pm 1, \pm 3
Factors of 2: \pm 1, \pm 2
All combinations for the rational zeros can be found by using just the positive factors of the leading coefficient. The possible rational zeros are
\frac{\pm 1}{1}, \frac{\pm 1}{2}, \frac{\pm 3}{1}, \frac{\pm 3}{2}
or
\pm 1, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm 3 .
We use synthetic division to determine whether a rational zero exists among these eight candidates. We start by testing 1. If 1 is not a rational zero, we will test other possibilities.
\begin{matrix} \underline{1|} & \\ \\ \\ \\ \end{matrix} \begin{matrix} 2 & 5& -4 & -3 \\ & 2 & 7 & 3\\ \hline 2 &7 & 3 & |0 \end{matrix}
The zero remainder tells us that (x-1) is a factor of F(x), with the other factor being 2 x²+7 x+3. So
\begin{aligned}F(x) &=(x-1)\left(2 x^{2}+7 x+3\right) \\&=(x-1)(2 x+1)(x+3) \quad \text { Factor } 2 x^{2}+7 x+3\end{aligned}
We find the zeros of F(x) by solving the equation (x-1)(2 x+1)(x+3)=0.
x-1=0 \quad \text { or } 2 x+1=0 \text { or } x+3=0 Zero-product property
x=1 \quad \text { or } \quad x=-\frac{1}{2} \quad \text { or } \quad x=-3 Solve each equation.
The solution set is \left\{1,-\frac{1}{2},-3\right\}. The rational zeros of F are -3,-\frac{1}{2}, \text { and } 1.