Use the Remainder Theorem to Completely Factor P(x) = x³ – 6x² + 11x – 6

The Remainder Theorem states that if you divide a polynomial P(x) by a linear divisor (x – c), the remainder of that division is equal to P(c). To completely factor the polynomial P(x) = x³ – 6x² + 11x – 6, we first need to find its possible rational roots.

Using the Rational Root Theorem, the potential rational roots are the factors of the constant term, which is -6. The possible rational roots are ±1, ±2, ±3, ±6.

We will test these values in P(x) to find a root. Let’s start with x = 1:

P(1) = 1³ - 6(1)² + 11(1) - 6 = 1 - 6 + 11 - 6 = 0

Since P(1) = 0, x – 1 is a factor of P(x). Now, we can divide P(x) by (x – 1) using synthetic division:

1 | 1  -6  11  -6
  |     1  -5   6
_________________
    1  -5   6   0

The result from synthetic division gives us the quotient x² – 5x + 6 with a remainder of 0. Now we have:

P(x) = (x - 1)(x² - 5x + 6)

Next, we need to factor the quadratic x² – 5x + 6. We look for two numbers that multiply to 6 and add to -5. The numbers -2 and -3 fit this requirement. Therefore:

x² - 5x + 6 = (x - 2)(x - 3)

Combining it all, we can rewrite P(x) as:

P(x) = (x - 1)(x - 2)(x - 3)

Thus, the complete factorization of P(x) = x³ – 6x² + 11x – 6 using the Remainder Theorem is:

P(x) = (x - 1)(x - 2)(x - 3)