Problem 177 - College Algebra

Question

Exercises \(177-179\) will help you prepare for the material covered in the next section. Factor completely: \(x^{3}+x^{2}-4 x-4\)

Step-by-Step Solution

Answer

The factorized form of the given cubic equation is \((x-2)(x+2)(x+1)\

Step 1: Grouping terms

The first step is to group the terms. Lets take \(x^{3}+x^{2}\) and \(-4 x-4\). So the equation now reads \((x^{3}+x^{2})-4(x+1)\)

Step 2: Factoring the common terms

Within each group, we look for common factors. We factor out \(x^2\) from the first group and -4 from the second group, leaving us with: \(x^{2}(x+1)-4(x+1)\)

Step 3: Factor the results

Both terms in the equation now share a common factor of \(x+1\). We factorise the \(x+1\) from the equation which gives us \((x^2-4)(x+1)\)

Step 4: Factoring difference of squares

Notice that \(x^2 - 4\) is a difference of squares which can be factored as \((x-2)(x+2)\), giving the final expression as \((x-2)(x+2)(x+1)\)

Key Concepts

Grouping Terms

Grouping terms is an essential strategy in factoring more complex polynomials. This method involves pairing terms in the polynomial in such a way that makes it easier to factor them down. In exercises like these, an expression like \(x^3 + x^2 - 4x - 4\) can initially seem daunting. However, if we carefully look at the polynomial, we can break it into parts to simplify our task.

• First, identify terms that can be grouped together. For instance, group \(x^3 + x^2\) and \(-4x - 4\).
• Next, write these groups separately in parentheses: \((x^3 + x^2)\) and \((-4x - 4)\).

This initial regrouping sets up a pathway for easier factorization, making the polynomial seem less complex by dealing with each group individually.

Common Factors

Once you have grouped the terms, the next step is identifying any common factors within these groups. A common factor is a number or variable that divides all terms in the group without a remainder.
For \(x^3 + x^2\), \(x^2\) is a common factor as it can be taken out of both \(x^3\) and \(x^2\). Thus, this group becomes \(x^2(x + 1)\).

• This works similarly for the second group: \(-4x - 4\). Here, \(-4\) is a common factor. Factoring \(-4\) out gives us \(-4(x + 1)\).

The key is to look for terms that appear in every part of the group. By factoring these out, you simplify the expression considerably and set the stage for even more efficient factorization.

Difference of Squares

The concept of the difference of squares is a powerful technique that helps in further breaking down an expression into simpler factors. This identity states that any expression of the form \(a^2 - b^2\) can be factored as \((a - b)(a + b)\).
In our polynomial, after taking common factors out, we are left with \((x^2 - 4)(x + 1)\). Here, \(x^2 - 4\) fits the pattern of a difference of squares because it can be rewritten as \(x^2 - 2^2\). Applying the identity:

• This becomes \((x - 2)(x + 2)\).

By recognizing and applying the difference of squares pattern, you achieve the simplest form of the polynomial. Therefore, the complete factorization results in \((x-2)(x+2)(x+1)\), showcasing the ease and power of this technique.