Question

Decompose the polynomial into linear factors: (x2−7x+6)(x2+3x−18)
(1 point)
Responses

(x−7)(x+6)(x+3)(x−18)
open paren x minus 7 close paren times open paren x plus 6 close paren times open paren x plus 3 close paren times open paren x minus 18 close paren

(x−6)(x−1)(x+6)(x−3)
open paren x minus 6 close paren times open paren x minus 1 close paren times open paren x plus 6 close paren times open paren x minus 3 close paren

(x−3)(x−2)(x+6)(x−3)
open paren x minus 3 close paren times open paren x minus 2 close paren times open paren x plus 6 close paren times open paren x minus 3 close paren

(x−6)(x+1)(x−9)(x−2)

Answer 1

To decompose the polynomial \((x^2 − 7x + 6)(x^2 + 3x − 18)\) into linear factors, we first need to factor each quadratic expression separately.

1. Factor \(x^2 - 7x + 6\):
- We are looking for two numbers that multiply to \(6\) (the constant term) and add to \(-7\) (the coefficient of \(x\)).
- The numbers are \(-1\) and \(-6\).
- Thus, \(x^2 - 7x + 6 = (x - 1)(x - 6)\).

2. Factor \(x^2 + 3x - 18\):
- We are looking for two numbers that multiply to \(-18\) and add to \(3\).
- The numbers are \(6\) and \(-3\).
- Thus, \(x^2 + 3x - 18 = (x + 6)(x - 3)\).

Now we can combine the factored forms:
\[
(x^2 - 7x + 6)(x^2 + 3x - 18) = (x - 1)(x - 6)(x + 6)(x - 3)
\]

Putting all together, the final linear factored form of the polynomial is:
\[
(x - 1)(x - 6)(x + 6)(x - 3)
\]

However, the answer options provided do not exactly match this expression. The closest option appears to be:

**(x−6)(x−1)(x+6)(x−3)**

So, the correct option is:
\[
\text{(x−6)(x−1)(x+6)(x−3)}
\]