Question

solve the given polynomial equation. use the rational zero theorem and descartess rule of signs as an aid in obtaining the first root. 2x^3 - 5x^2 - 5x - 1 = 0 the solution set is {}. (use commas to separate answers. type integers or fractions. type exact answers using radicals as needed.)

Explanation:

Step1: Apply Rational Zero Theorem

The Rational Zero Theorem states that if a polynomial equation $a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0=0$ has integer coefficients, then the possible rational zeros are of the form $\frac{p}{q}$, where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading - coefficient $a_n$. For the polynomial $2x^{3}-5x^{2}-5x - 1 = 0$, $a_n = 2$ and $a_0=-1$. The factors of $a_0=-1$ are $\pm1$, and the factors of $a_n = 2$ are $\pm1,\pm2$. So the possible rational zeros are $\pm1,\pm\frac{1}{2}$.

Step2: Test possible rational zeros

Let $f(x)=2x^{3}-5x^{2}-5x - 1$.
$f(1)=2(1)^{3}-5(1)^{2}-5(1)-1=2 - 5-5 - 1=-9$.
$f(-1)=2(-1)^{3}-5(-1)^{2}-5(-1)-1=-2 - 5 + 5-1=-3$.
$f(\frac{1}{2})=2(\frac{1}{2})^{3}-5(\frac{1}{2})^{2}-5(\frac{1}{2})-1=2\times\frac{1}{8}-5\times\frac{1}{4}-\frac{5}{2}-1=\frac{1}{4}-\frac{5}{4}-\frac{10}{4}-\frac{4}{4}=\frac{1 - 5-10 - 4}{4}=-\frac{18}{4}=-\frac{9}{2}$.
$f(-\frac{1}{2})=2(-\frac{1}{2})^{3}-5(-\frac{1}{2})^{2}-5(-\frac{1}{2})-1=2\times(-\frac{1}{8})-5\times\frac{1}{4}+\frac{5}{2}-1=-\frac{1}{4}-\frac{5}{4}+\frac{10}{4}-\frac{4}{4}=0$. So $x =-\frac{1}{2}$ is a root.

Step3: Perform polynomial long - division

Divide $2x^{3}-5x^{2}-5x - 1$ by $(2x + 1)$.
Using polynomial long - division:
\[

$$\begin{align*} \frac{2x^{3}-5x^{2}-5x - 1}{2x+1}&=x^{2}-3x - 1 \end{align*}$$

\]

Step4: Solve the quadratic equation

We now solve the quadratic equation $x^{2}-3x - 1=0$.
The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
For $x^{2}-3x - 1=0$, $a = 1$, $b=-3$, $c=-1$.
$x=\frac{3\pm\sqrt{(-3)^{2}-4(1)(-1)}}{2(1)}=\frac{3\pm\sqrt{9 + 4}}{2}=\frac{3\pm\sqrt{13}}{2}$.

Answer:

$-\frac{1}{2},\frac{3+\sqrt{13}}{2},\frac{3 - \sqrt{13}}{2}$