Answer :
To solve the polynomial equation [tex]\(4x^3 - 12x^2 - 9x - 1 = 0\)[/tex], we can use the Rational Zero Theorem and Descartes's Rule of Signs as aids in finding the first root.
### Step-by-Step Solution:
1. Rational Zero Theorem:
The Rational Zero Theorem states that any rational solution of the polynomial equation can be written in the form [tex]\(\frac{p}{q}\)[/tex], where [tex]\(p\)[/tex] is a factor of the constant term (in this case, -1) and [tex]\(q\)[/tex] is a factor of the leading coefficient (in this case, 4).
The possible rational roots are therefore:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4} \][/tex]
2. Descartes's Rule of Signs:
Descartes's Rule of Signs helps us determine the possible number of positive and negative roots by examining the signs of the coefficients.
- Positive roots: When we look at the polynomial [tex]\(4x^3 - 12x^2 - 9x - 1 = 0\)[/tex], we notice that there are two sign changes ([tex]\( + \rightarrow - \text{ and } - \rightarrow -\)[/tex]) indicating either 2 or 0 positive roots.
- Negative roots: For negative roots, we substitute [tex]\(x\)[/tex] with [tex]\(-x\)[/tex] and get the polynomial [tex]\(4(-x)^3 - 12(-x)^2 - 9(-x) - 1 = -4x^3 - 12x^2 + 9x - 1\)[/tex]. There are three sign changes ([tex]\(-\rightarrow - \rightarrow + \rightarrow -\)[/tex]), indicating 3 or 1 negative roots.
3. Testing the Possible Rational Roots:
We substitute each possible rational root into the polynomial to see which one makes the equation equal to zero.
Testing [tex]\(x = -\frac{1}{2}\)[/tex]:
[tex]\[ 4\left(-\frac{1}{2}\right)^3 - 12\left(-\frac{1}{2}\right)^2 - 9\left(-\frac{1}{2}\right) - 1 = -\frac{1}{2} + 3 + \frac{9}{2} - 1 = 0 \][/tex]
[tex]\(x = -\frac{1}{2}\)[/tex] is a root.
4. Factor out [tex]\( (x + \frac{1}{2}) \)[/tex] using Polynomial Division:
Now that we know [tex]\(x = -\frac{1}{2}\)[/tex] is a root, we can factor [tex]\((x + \frac{1}{2})\)[/tex] out of the polynomial. We divide the polynomial [tex]\(4x^3 - 12x^2 - 9x - 1\)[/tex] by [tex]\((x + \frac{1}{2})\)[/tex] to get the quotient polynomial.
The quotient polynomial is:
[tex]\[ 4x^2 - 14x - 2 = 0 \][/tex]
5. Solving the Quadratic Polynomial:
We solve the quadratic polynomial [tex]\(4x^2 - 14x - 2 = 0\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 4\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = -2\)[/tex].
[tex]\[ x = \frac{14 \pm \sqrt{196 + 32}}{8} = \frac{14 \pm \sqrt{228}}{8} = \frac{14 \pm 2\sqrt{57}}{8} = \frac{7 \pm \sqrt{57}}{4} \][/tex]
### Final Solution Set:
The exact roots of the polynomial equation [tex]\(4x^3 - 12x^2 - 9x - 1 = 0\)[/tex] are:
[tex]\[ \left\{-\frac{1}{2}, \frac{7 - \sqrt{57}}{4}, \frac{7 + \sqrt{57}}{4}\right\} \][/tex]
### Step-by-Step Solution:
1. Rational Zero Theorem:
The Rational Zero Theorem states that any rational solution of the polynomial equation can be written in the form [tex]\(\frac{p}{q}\)[/tex], where [tex]\(p\)[/tex] is a factor of the constant term (in this case, -1) and [tex]\(q\)[/tex] is a factor of the leading coefficient (in this case, 4).
The possible rational roots are therefore:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4} \][/tex]
2. Descartes's Rule of Signs:
Descartes's Rule of Signs helps us determine the possible number of positive and negative roots by examining the signs of the coefficients.
- Positive roots: When we look at the polynomial [tex]\(4x^3 - 12x^2 - 9x - 1 = 0\)[/tex], we notice that there are two sign changes ([tex]\( + \rightarrow - \text{ and } - \rightarrow -\)[/tex]) indicating either 2 or 0 positive roots.
- Negative roots: For negative roots, we substitute [tex]\(x\)[/tex] with [tex]\(-x\)[/tex] and get the polynomial [tex]\(4(-x)^3 - 12(-x)^2 - 9(-x) - 1 = -4x^3 - 12x^2 + 9x - 1\)[/tex]. There are three sign changes ([tex]\(-\rightarrow - \rightarrow + \rightarrow -\)[/tex]), indicating 3 or 1 negative roots.
3. Testing the Possible Rational Roots:
We substitute each possible rational root into the polynomial to see which one makes the equation equal to zero.
Testing [tex]\(x = -\frac{1}{2}\)[/tex]:
[tex]\[ 4\left(-\frac{1}{2}\right)^3 - 12\left(-\frac{1}{2}\right)^2 - 9\left(-\frac{1}{2}\right) - 1 = -\frac{1}{2} + 3 + \frac{9}{2} - 1 = 0 \][/tex]
[tex]\(x = -\frac{1}{2}\)[/tex] is a root.
4. Factor out [tex]\( (x + \frac{1}{2}) \)[/tex] using Polynomial Division:
Now that we know [tex]\(x = -\frac{1}{2}\)[/tex] is a root, we can factor [tex]\((x + \frac{1}{2})\)[/tex] out of the polynomial. We divide the polynomial [tex]\(4x^3 - 12x^2 - 9x - 1\)[/tex] by [tex]\((x + \frac{1}{2})\)[/tex] to get the quotient polynomial.
The quotient polynomial is:
[tex]\[ 4x^2 - 14x - 2 = 0 \][/tex]
5. Solving the Quadratic Polynomial:
We solve the quadratic polynomial [tex]\(4x^2 - 14x - 2 = 0\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 4\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = -2\)[/tex].
[tex]\[ x = \frac{14 \pm \sqrt{196 + 32}}{8} = \frac{14 \pm \sqrt{228}}{8} = \frac{14 \pm 2\sqrt{57}}{8} = \frac{7 \pm \sqrt{57}}{4} \][/tex]
### Final Solution Set:
The exact roots of the polynomial equation [tex]\(4x^3 - 12x^2 - 9x - 1 = 0\)[/tex] are:
[tex]\[ \left\{-\frac{1}{2}, \frac{7 - \sqrt{57}}{4}, \frac{7 + \sqrt{57}}{4}\right\} \][/tex]
