Question

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:
\[
\boxed{1, 2, 3}
\]