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 \(p=\pm1\), and the factors of \(a_n = 2\) are \(q=\pm1,\pm2\). So the possible rational zeros are \(\pm1,\pm\frac{1}{2}\).

Step2: Test possible rational zeros

Evaluate the polynomial at the possible rational zeros:

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

Step3: Perform polynomial long - division

Since \(x =-\frac{1}{2}\) is a root, \((2x + 1)\) is a factor of \(2x^{3}-5x^{2}-5x - 1\). Divide \(2x^{3}-5x^{2}-5x - 1\) by \(2x + 1\) using polynomial long - division or synthetic division. Using synthetic division with \(c=-\frac{1}{2}\):

- \(\frac{1}{2}\)	2	- 5	- 5	- 1
	2	- 6	- 2	0

The quotient is \(x^{2}-3x - 1\).

Step4: Solve the quadratic equation

Set \(x^{2}-3x - 1=0\). Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b=-3\), and \(c=-1\). Then \(x=\frac{3\pm\sqrt{(-3)^{2}-4(1)(-1)}}{2(1)}=\frac{3\pm\sqrt{9 + 4}}{2}=\frac{3\pm\sqrt{13}}{2}\).

Answer:

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