The factors of the polynomial x3 - 6x2 + 11x - 6 are:

Concept

For a polynomial P(x), the Factor Theorem states that (x - r) is a factor exactly when P(r) = 0.

For a monic cubic with roots a, b and c, P(x) = (x - a)(x - b)(x - c). Its coefficients can be checked using the sum and products of the roots.

Application

1. Evaluate P(1): 1 - 6 + 11 - 6 = 0. Therefore, (x - 1) is a factor.

2. Divide P(x) by (x - 1). The quotient is x2 - 5x + 6.

3. Find two numbers whose sum is 5 and product is 6. They are 2 and 3, so x2 - 5x + 6 = (x - 2)(x - 3).

4. Hence, P(x) = (x - 1)(x - 2)(x - 3).

Cross-check

1. First, (x - 2)(x - 3) = x2 - 5x + 6.

2. Then, (x - 1)(x2 - 5x + 6) = x3 - 6x2 + 11x - 6, which reproduces the given polynomial.

Result: The factorization is (x - 1)(x - 2)(x - 3).