Question

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime. $$ x^{3}+2 x^{2}-x-2 $$

Short answer

(x + 2)(x + 1)(x - 1)

Step-by-step solution

- Group the terms

Group the polynomial into two pairs: \( (x^3 + 2x^2) + (-x - 2) \)

- Factor out the Greatest Common Factor (GCF) from each pair

Factor out the GCF from each group: \[ x^2(x+2) - 1(x+2) \]

- Factor by grouping

Notice that \( x + 2 \) is common in both terms. Factor \( x+2 \) out: \[ (x+2)(x^2 - 1) \]

- Factor the difference of squares

\( x^2 - 1 \) can be factored further using the difference of squares formula: \[ (x + 2)(x + 1)(x - 1) \]

Key concepts

Grouping Terms

Grouping terms is a technique used to simplify polynomials. When you have a polynomial with four terms, you can attempt to group them into pairs to make factoring easier.

Let's start with the polynomial: \[ x^3 + 2x^2 - x - 2 \]

Here, you place the first two terms in one group and the last two in another:

\[ (x^3 + 2x^2) + (-x - 2) \]

At this point, each group can be handled separately to make further factoring easier. Grouping terms helps in breaking down the polynomial into simpler parts that can be addressed individually.

Greatest Common Factor

The Greatest Common Factor (GCF) is the largest factor that divides two numbers. For polynomials, it extends to common variables and coefficients. Identifying the GCF simplifies the process of factoring each group.

For the grouped pairs in our example:

\[ (x^3 + 2x^2) + (-x - 2) \]

We can see that the GCF of the first group \((x^3 + 2x^2)\) is \(x^2\). Factoring it out gives:

\[ x^2(x + 2) \]

For the second group \((-x - 2)\), it is -1. Factoring this out we get:

\[ -1(x + 2) \]

So after factoring the GCF from both groups, the polynomial looks like this:

\[ x^2(x + 2) - 1(x + 2) \]

Factoring by Grouping

After factoring out the GCF from the grouped terms, check if each term contains a common binomial factor. In our example:

\[ x^2(x + 2) - 1(x + 2) \]

Both terms include \((x + 2)\), allowing us to factor it out as a common factor:

\[ (x + 2)(x^2 - 1) \]

Factoring by grouping leverages the common factor in each term to simplify the polynomial further. It's a useful technique to reduce the polynomial into a product of simpler expressions.

Difference of Squares

The difference of squares is a specific factoring pattern where a binomial is written as the difference of two squares. It's represented by the formula:\

\( a^2 - b^2 = (a - b)(a + b)\)

In the polynomial \( x^2 - 1 \), recognize that it fits the difference of squares formula:\

\( x^2 - 1 = (x)^2 - (1)^2 \).

This can be factored into:\

\((x - 1)(x + 1)\).

So, our factored polynomial finally is:

\((x + 2)(x - 1)(x + 1)\).

Using the difference of squares technique allows you to break down complex expressions into simpler binomials quickly and effectively.