Find the roots of the polynomial: x³ - 6x² + 11x - 6 = 0.

Answer 1

To find the roots of the polynomial \( x^3 - 6x^2 + 11x - 6 = 0 \), we can start by using the Rational Root Theorem to test possible rational roots.

The possible rational roots of the polynomial are the factors of the constant term (-6) divided by the factors of the leading coefficient (1). So the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6 \).

Let's evaluate the polynomial at these values:

1. Testing \( x = 1 \): \[ 1^3 - 6(1^2) + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 \] So, \( x = 1 \) is a root.

Now that we have a root, we can factor the polynomial by dividing it by \( x - 1 \). We will perform synthetic division with \( x = 1 \).

Setting up synthetic division:

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

The result of the synthetic division is \( x^2 - 5x + 6 \).

Now we will factor \( x^2 - 5x + 6 \): \[ x^2 - 5x + 6 = (x - 2)(x - 3) \]

Combining everything, we get: \[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) \]

Now we can find the roots: Setting each factor equal to zero:

1. \( x - 1 = 0 \) \( \Rightarrow x = 1 \)
2. \( x - 2 = 0 \) \( \Rightarrow x = 2 \)
3. \( x - 3 = 0 \) \( \Rightarrow x = 3 \)

Thus, the roots of the polynomial \( x^3 - 6x^2 + 11x - 6 = 0 \) are: \[ \boxed{1, 2, 3} \]