Problem 38 - Elementary Algebra

Question

Factor completely. $$ 4 a_{2} b_{2}-4 a_{2}-9 b_{2}+9 $$

Step-by-Step Solution

Answer

The completely factored form is \((2a - 3)(2a + 3)(b - 1)(b + 1)\).

Step 1: Arrange Terms

First, let's rearrange the expression given: \(4a^2b^2 - 4a^2 - 9b^2 + 9\). This will help us in identifying groups that we might be able to factor further.

Step 2: Group and Factor

Now, group the terms: \((4a^2b^2 - 4a^2) + (-9b^2 + 9)\). Factor out common factors from each group: \(4a^2(b^2 - 1) - 9(b^2 - 1)\).

Step 3: Factor by Grouping

Notice that \((b^2 - 1)\) is a common factor. This allows us to factor by grouping: \((4a^2 - 9)(b^2 - 1)\).

Step 4: Recognize Difference of Squares

Recognize each term is a difference of squares. Apply the difference of squares formula \(x^2 - y^2 = (x-y)(x+y)\).For \(4a^2 - 9\):It can be rewritten as \((2a)^2 - (3)^2\), which factors into \((2a - 3)(2a + 3)\).For \(b^2 - 1\):It can be rewritten as \(b^2 - 1^2\), which factors into \((b - 1)(b + 1)\).

Step 5: Write the Completely Factored Form

Combine the expressions from the previous step to write the completely factored form: \((2a - 3)(2a + 3)(b - 1)(b + 1)\).

Key Concepts

Factoring

Factoring is an essential algebraic technique used to simplify expressions or solve equations. When you factor an expression, you break it down into simpler multipliers that, when multiplied together, will give the original expression. This process is crucial for simplifying algebraic expressions and solving quadratic equations.

To factor an algebraic expression:

• First, look for a greatest common factor (GCF) in all the terms. If one exists, factor it out.
• Next, rearrange terms, if necessary, to identify common factors within groups of terms.
• Then, apply different factoring techniques, such as the difference of squares formula or the grouping method, to simplify further.

Factoring helps express complex equations in a more workable form, allowing for easier manipulation and solution finding. It’s important in algebra to master this technique, as it is foundational for understanding future mathematical concepts.

Difference of Squares

The difference of squares is a specific algebraic form: two squared terms being subtracted from each other. It’s recognized as one of the key identities in algebra and is given by the formula:\[ x^2 - y^2 = (x-y)(x+y) \] This formula is incredibly powerful because it allows you to quickly and easily factor expressions that fit the difference of squares pattern.

For example, in the expression from our exercise, both \(4a^2 - 9\) and \(b^2 - 1\) are differences of squares. Let's break them down:

• \(4a^2 - 9\) can be rewritten as \((2a)^2 - (3)^2\). Using the difference of squares formula, it factors to \((2a - 3)(2a + 3)\).
• Similarly, \(b^2 - 1\) can be seen as \(b^2 - 1^2\), which factors into \((b - 1)(b + 1)\).

Recognizing the difference of squares is a valuable skill as it makes factoring and solving problems smoother and more efficient.

Grouping Method

The grouping method is another powerful strategy used in algebra to factor polynomials. It is especially useful when dealing with four-term groupings, as it allows for systematic reduction into more manageable expressions.

Here's how the grouping method works:

• First, divide the polynomial into pairs or groups. Ideally, each group should have a common factor.
• Then, factor out the common factor from each group. This might change the appearance of the polynomial, making it easier to handle.
• Once common factors are factored out, look for a common binomial factor. This allows the expression to be rewritten, often simplifying into a product of binomials.

In our original exercise, we used the grouping method to factor the expression by first grouping the terms: \((4a^2b^2 - 4a^2) + (-9b^2 + 9)\). By factoring common elements from these groupings, we ended up with a common binomial, allowing us to further simplify the expression. Mastery of this method enhances problem-solving skills and deepens algebraic understanding.