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Chapter 6: Problem 2

Factor completely. Identify any prime polynomials. $$ 2 x^{2}-18 $$

### Short Answer

Expert verified

2(x+3)(x-3)

## Step by step solution

01

## Identify the Greatest Common Factor (GCF)

First, look for the greatest common factor (GCF) of the terms in the polynomial. The given polynomial is 2x^{2} - 18. The GCF of 2 and -18 is 2.

02

## Factor out the GCF

Next, factor out the GCF from each term in the polynomial. This means we divide each term by 2: 2(x^{2} - 9).

03

## Factor the Quadratic Expression

Now, factor the quadratic expression inside the parenthesis, x^{2} - 9. Notice that this is a difference of squares which can be factored as: (x+3)(x-3).

04

## Combine All Factors

Combine the GCF with the factored quadratic expression to get the completely factored form: 2(x+3)(x-3).

## Key Concepts

These are the key concepts you need to understand to accurately answer the question.

###### Greatest Common Factor (GCF)

The Greatest Common Factor (GCF) is the largest factor that can evenly divide all terms of a polynomial.

In our exercise, we have the polynomial \(2x^2 - 18\). We need to identify the GCF of the coefficients 2 and -18.

Since both numbers share the factor 2, it is the GCF.

Factoring out the GCF simplifies the polynomial by removing this factor from each term. For our polynomial, it turns into \(2(x^2 - 9)\).

This step is crucial because it makes the next steps of factoring much easier.

###### Difference of Squares

The difference of squares is a specific type of quadratic expression. It takes the form \(a^2 - b^2\).

This can always be factored into \((a+b)(a-b)\).

In our exercise, we reach the expression \(x^2 - 9\) after factoring out the GCF. Notice that this can be rewritten as \(x^2 - 3^2\), fitting the difference of squares format.

Applying the difference of squares rule, \(x^2 - 9\) becomes \((x+3)(x-3)\). This step exploits the structure of certain quadratics and offers a straightforward factoring method.

###### Quadratic Expressions

Quadratic expressions are polynomials that follow the form \(ax^2 + bx + c\). In many cases, they can be factored into products of linear binomials.

In our problem, once we factor out the GCF, we are left with \(x^2 - 9\), a simple quadratic with no linear term. This is a special case since it's also a difference of squares.

Generally, the approach involves: identifying patterns (like the difference of squares), factoring out the GCF first, and then breaking down the quadratic expression further.

For more complex quadratics, formulas such as the quadratic formula or completing the square might be used.

But in our exercise, recognizing the difference of squares allowed a simpler solution.

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