To solve the equation [tex]\(x^3 - 6x^2 + 11x - 6 = 0\)[/tex], we will look for the roots of the polynomial. Here's a detailed, step-by-step breakdown of the solution:
1. First, note that the polynomial given is [tex]\(x^3 - 6x^2 + 11x - 6\)[/tex].
2. To find the roots of this polynomial, we need to solve [tex]\(x^3 - 6x^2 + 11x - 6 = 0\)[/tex].
3. One method of solving cubic equations involves finding the potential rational roots using the Rational Root Theorem. However, instead, we realize that the polynomial can be factored.
4. We aim to factor this polynomial into simpler components:
[tex]\[
x^3 - 6x^2 + 11x - 6 = (x - a_1)(x - a_2)(x - a_3)
\][/tex]
where [tex]\(a_1\)[/tex], [tex]\(a_2\)[/tex], and [tex]\(a_3\)[/tex] are the roots of the polynomial.
5. To factor this cubic equation:
- Consider the roots as likely integers since it's always straightforward to check integer potential roots like [tex]\( \pm 1, \pm 2, \pm 3, ...\)[/tex].
6. In trial method, observing the polynomial and via professional experience, we efficiently determine the roots are:
- [tex]\(x = 1\)[/tex]
- [tex]\(x = 2\)[/tex]
- [tex]\(x = 3\)[/tex]
7. To validate, substitute these roots back into the polynomial:
- For [tex]\(x = 1\)[/tex]:
[tex]\[
1^3 - 6 \cdot 1^2 + 11 \cdot 1 - 6 = 1 - 6 + 11 - 6 = 0
\][/tex]
- For [tex]\(x = 2\)[/tex]:
[tex]\[
2^3 - 6 \cdot 2^2 + 11 \cdot 2 - 6 = 8 - 24 + 22 - 6 = 0
\][/tex]
- For [tex]\(x = 3\)[/tex]:
[tex]\[
3^3 - 6 \cdot 3^2 + 11 \cdot 3 - 6 = 27 - 54 + 33 - 6 = 0
\][/tex]
Each substitution confirms that the roots satisfy the polynomial equation.
8. As a result, we determine that the roots of the polynomial [tex]\(x^3 - 6x^2 + 11x - 6 = 0\)[/tex] are:
[tex]\[
\boxed{1, 2, 3}
\][/tex]
Therefore, the set [tex]\(B\)[/tex] which consists of all solutions to the equation is:
[tex]\[
B = \{1, 2, 3\}
\][/tex]
