Question

Decompose the polynomial into linear factors (x^2 - 7x + 6) (x^2 + 3x -18)

.

A
(x−6)(x−1)(x+6)(x+3)\left(x-6\right)\left(x-1\right)\left(x+6\right)\left(x+3\right)(x−6)(x−1)(x+6)(x+3)

B
(x+6)(x+1)(x−6)(x+3)\left(x+6\right)\left(x+1\right)\left(x-6\right)\left(x+3\right)(x+6)(x+1)(x−6)(x+3)

C
(x−6)(x−1)(x+6)(x−3)\left(x-6\right)\left(x-1\right)\left(x+6\right)\left(x-3\right)(x−6)(x−1)(x+6)(x−3)

D
(x−6)(x+1)(x−3)(x−6)\left(x-6\right)\left(x+1\right)\left(x-3\right)\left(x-6\right)(x−6)(x+1)(x−3)(x−6)

Answer 1

To decompose the polynomial \((x^2 - 7x + 6)(x^2 + 3x - 18)\) into linear factors, we will begin by factoring each quadratic polynomial individually.

1. **Factor \(x^2 - 7x + 6\)**:
We need two numbers that multiply to \(6\) (the constant term) and add to \(-7\) (the coefficient of \(x\)):
\[
x^2 - 7x + 6 = (x - 6)(x - 1)
\]

2. **Factor \(x^2 + 3x - 18\)**:
We need two numbers that multiply to \(-18\) and add to \(3\):
\[
x^2 + 3x - 18 = (x + 6)(x - 3)
\]

Now, we combine these factorizations together:
\[
(x^2 - 7x + 6)(x^2 + 3x - 18) = (x - 6)(x - 1)(x + 6)(x - 3)
\]

Thus, the complete factorization of the polynomial is:
\[
(x - 6)(x - 1)(x + 6)(x - 3)
\]

Now, let's check the answer options:
- Option A: \((x-6)(x-1)(x+6)(x+3)\)
- Option B: \((x+6)(x+1)(x−6)(x+3)\)
- Option C: \((x−6)(x−1)(x+6)(x−3)\)
- Option D: \((x−6)(x+1)(x−3)(x−6)\)

The correct choice is:
**C**: \((x−6)(x−1)(x+6)(x−3)\)