To solve the polynomial equation [tex]\( 8x^3 - 2x^2 - 5x - 1 = 0 \)[/tex], let's consider its solutions step-by-step.
1. Identify the polynomial: We start with the polynomial
[tex]\[ 8x^3 - 2x^2 - 5x - 1 = 0 \][/tex]
2. Look for rational roots: One way to begin solving polynomial equations is to use the Rational Root Theorem. According to the theorem, any rational root, if it exists, must be a factor of the constant term (-1) divided by a factor of the leading coefficient (8).
- Possible factors of [tex]\(-1\)[/tex] are [tex]\(\pm 1\)[/tex].
- Possible factors of [tex]\(8\)[/tex] are [tex]\(\pm 1, \pm 2, \pm 4, \pm 8\)[/tex].
Hence, the potential rational roots are:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8} \][/tex]
3. Testing potential roots: We test these potential roots to see if they are solutions to the polynomial equation.
After testing, we find the roots are:
[tex]\[ x = -\frac{1}{2}, -\frac{1}{4}, 1 \][/tex]
4. Verify the roots: To ensure these are indeed the roots, we can substitute them back into the original equation:
- For [tex]\( x = -\frac{1}{2} \)[/tex]:
[tex]\[
8 \left( -\frac{1}{2} \right)^3 - 2 \left( -\frac{1}{2} \right)^2 - 5 \left( -\frac{1}{2} \right) - 1
= 8 \left( -\frac{1}{8} \right) - 2 \left( \frac{1}{4} \right) + \frac{5}{2} - 1
= -1 - \frac{1}{2} + \frac{5}{2} - 1
= 0
\][/tex]
- For [tex]\( x = -\frac{1}{4} \)[/tex]:
[tex]\[
8 \left( -\frac{1}{4} \right)^3 - 2 \left( -\frac{1}{4} \right)^2 - 5 \left( -\frac{1}{4} \right) - 1
= 8 \left( -\frac{1}{64} \right) - 2 \left( \frac{1}{16} \right) + \frac{5}{4} - 1
= -\frac{1}{8} - \frac{1}{8} + \frac{5}{4} - 1
= 0
\][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[
8(1)^3 - 2(1)^2 - 5(1) - 1
= 8 - 2 - 5 - 1
= 0
\][/tex]
Thus, the verified solutions to the equation [tex]\( 8x^3 - 2x^2 - 5x - 1 = 0 \)[/tex] are:
[tex]\[ x = -\frac{1}{2}, -\frac{1}{4}, 1 \][/tex]
So, the roots of the polynomial equation are:
[tex]\[ x = -\frac{1}{2}, -\frac{1}{4}, 1 \][/tex]
