Question

Write the product as linear factors: (x2−4)(x2−9)

Answer 1

Decompose the polynomial x3−2x2−5x+6
into its linear factors, given that x−1
is a factor.

Answer 2

To express the product \((x^2 - 4)(x^2 - 9)\) as linear factors, we can start by factoring each quadratic expression individually.

1. The first factor \(x^2 - 4\) is a difference of squares:
\[
x^2 - 4 = (x - 2)(x + 2)
\]

2. The second factor \(x^2 - 9\) is also a difference of squares:
\[
x^2 - 9 = (x - 3)(x + 3)
\]

Now, we can substitute these factorizations back into the product:
\[
(x^2 - 4)(x^2 - 9) = [(x - 2)(x + 2)][(x - 3)(x + 3)]
\]

Now, we can write the complete factorization as:
\[
(x - 2)(x + 2)(x - 3)(x + 3)
\]

Thus, the expression \((x^2 - 4)(x^2 - 9)\) as linear factors is:
\[
(x - 2)(x + 2)(x - 3)(x + 3)
\]

Answer 3

To decompose the polynomial \(x^3 - 2x^2 - 5x + 6\) into its linear factors, given that \(x - 1\) is a factor, we can use polynomial long division or synthetic division.

**Step 1: Perform synthetic division.**

We will divide \(x^3 - 2x^2 - 5x + 6\) by \(x - 1\) (which means we will use \(1\) in synthetic division).

```
1 | 1 -2 -5 6
| 1 -1 -6
---------------------
1 -1 -6 0
```

The result of the division gives us:
- The coefficients of the quotient are \(1\), \(-1\), and \(-6\). Therefore, the quotient is \(x^2 - x - 6\), and the remainder is \(0\).

This means:
\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6)
\]

**Step 2: Factor the quadratic \(x^2 - x - 6\).**

Next, we need to factor \(x^2 - x - 6\). We are looking for two numbers that multiply to \(-6\) and add to \(-1\). These numbers are \(-3\) and \(2\).

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

**Step 3: Write the complete factorization.**

Now we can combine everything:
\[
x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2)
\]

So, the polynomial \(x^3 - 2x^2 - 5x + 6\) decomposed into its linear factors is:
\[
(x - 1)(x - 3)(x + 2)
\]