Question

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

Answer 1

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

**Step 1: Factor \(x^2 - 7x + 6\)**

We look for two numbers that multiply to \(6\) (the constant term) and add to \(-7\) (the coefficient of \(x\)). The numbers \(-1\) and \(-6\) satisfy this because:
\[
-1 \times -6 = 6 \quad \text{and} \quad -1 + -6 = -7
\]

Thus, we can factor \(x^2 - 7x + 6\) as:
\[
(x - 1)(x - 6)
\]

**Step 2: Factor \(x^2 + 3x - 18\)**

Now we look for two numbers that multiply to \(-18\) (the constant term) and add to \(3\) (the coefficient of \(x\)). The numbers \(6\) and \(-3\) satisfy this because:
\[
6 \times -3 = -18 \quad \text{and} \quad 6 + -3 = 3
\]

Thus, we can factor \(x^2 + 3x - 18\) as:
\[
(x + 6)(x - 3)
\]

**Step 3: Combine the factors**

Now, we can combine all these factors together:

\[
(x^2 - 7x + 6)(x^2 + 3x - 18) = (x - 1)(x - 6)(x + 6)(x - 3)
\]

Therefore, the polynomial \((x^2 - 7x + 6)(x^2 + 3x - 18)\) can be decomposed into its linear factors as:
\[
(x - 1)(x - 6)(x + 6)(x - 3)
\]