How can I use the remainder theorem to completely factor the polynomial p(x) = x^3 - 6x^2 + 11x - 6?

Using the Remainder Theorem to Factor p(x)

To factor the polynomial p(x) = x^3 – 6x^2 + 11x – 6 using the remainder theorem, we first need to find potential rational roots. The remainder theorem states that if a polynomial f(x) is divided by x – c, the remainder of this division is equal to f(c). If f(c) = 0, then c is a root of the polynomial.

Step 1: Identify Potential Rational Roots

According to the Rational Root Theorem, possible rational roots can be found by taking the factors of the constant term (which is -6) and dividing them by the factors of the leading coefficient (which is 1). The factors of -6 are:

Step 2: Test Potential Roots

We will test these potential roots in the polynomial. Let’s start with x = 1:

p(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0

Since p(1) = 0, x = 1 is a root.

Step 3: Factor Out the Root

Having confirmed that 1 is a root, we can use synthetic division to divide p(x) by x – 1:

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

The result from synthetic division gives us:

p(x) = (x - 1)(x^2 - 5x + 6)

Step 4: Factor the Quadratic Expression

Next, we need to factor the quadratic x^2 – 5x + 6. This can be factored into:

x^2 - 5x + 6 = (x - 2)(x - 3)

Final Factorization

Thus, the complete factorization of the polynomial is:

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

In conclusion, we have successfully factored the polynomial p(x) = x^3 – 6x^2 + 11x – 6 completely as:

How can I use the remainder theorem to completely factor the polynomial p(x) = x^3 – 6x^2 + 11x – 6?