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

Answer 1

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 2

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

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: \[ (x - 1)(x - 3)(x + 2) \]