Decompose the polynomial into linear factors: (x^2 -5x+6)(x^2 +4x-12)

Question

1)(x-2)^2 (x-3)(x+2)
2)(x-1)(x-6)(x+2)(x-2)
3)(x-2)^2 (x-3)(x+6)
4)(x-2)(x-3)(x+2)(x-6)

Answer 1

To decompose the polynomial \( (x^2 - 5x + 6)(x^2 + 4x - 12) \) into linear factors, we can start by factoring each quadratic expression separately.

First, let's factor \( x^2 - 5x + 6 \): \[ x^2 - 5x + 6 = (x - 2)(x - 3) \] This is because the roots are \( x = 2 \) and \( x = 3 \).

Next, let's factor \( x^2 + 4x - 12 \): \[ x^2 + 4x - 12 = (x + 6)(x - 2) \] This is because the roots are \( x = -6 \) and \( x = 2 \).

Now we have: \[ (x^2 - 5x + 6)(x^2 + 4x - 12) = (x - 2)(x - 3)(x + 6)(x - 2) \] Notice that \( (x - 2) \) appears twice.

Combining all the factors, we get: \[ (x - 2)^2 (x - 3)(x + 6) \] Now we can match this with the given options:

1. \( (x - 2)^2 (x - 3)(x + 2) \)
2. \( (x - 1)(x - 6)(x + 2)(x - 2) \)
3. \( (x - 2)^2 (x - 3)(x + 6) \)
4. \( (x - 2)(x - 3)(x + 2)(x - 6) \)

The correct factorization is: 3) \( (x - 2)^2 (x - 3)(x + 6) \).