Problem 26 - College Algebra

Question

For the following exercises, factor the polynomial. $$ 4 m^{2}-9 $$

Step-by-Step Solution

Answer

The polynomial \(4m^2 - 9\) factors to \((2m - 3)(2m + 3)\).

Step 1: Identify the Form of the Polynomial

The given polynomial is \(4m^2 - 9\). Notice that it is a difference of squares. A difference of squares takes the form \(a^2 - b^2\), where \(a\) and \(b\) are real numbers.

Step 2: Express Each Term as a Square

Recognize the squares in the expression. The first term \(4m^2\) can be rewritten as \((2m)^2\), and the second term \(9\) can be rewritten as \((3)^2\). Thus, \(4m^2 - 9 = (2m)^2 - (3)^2\).

Step 3: Apply the Difference of Squares Formula

The difference of squares formula states that \(a^2 - b^2 = (a-b)(a+b)\). Here, \(a = 2m\) and \(b = 3\). Substituting these into the formula gives \((2m - 3)(2m + 3)\).

Step 4: Write the Final Factored Expression

Now that we have applied the formula, the polynomial \(4m^2-9\) factors to \((2m - 3)(2m + 3)\).

Key Concepts

Difference of Squares

The difference of squares is a unique mathematical identity that helps in factoring specific types of polynomial expressions. It applies to expressions that can be represented in the form \(a^2 - b^2\), where both \(a\) and \(b\) are real numbers or algebraic expressions. This structure highlights a special property: it can always be factored into two linear binomials, \((a-b)(a+b)\).

Here's how it works: the expression \(a^2 - b^2\) is changed into two separate terms by subtracting and adding the same value, \(b\). By doing this, a product structure is created that simplifies the expression.

• Example: \(x^2 - 16\) is the difference between \(x^2\) and \(4^2\). You factor it as \((x - 4)(x + 4)\).
• Example: \(9m^2 - 1\) becomes \((3m)^2 - (1)^2\), which factors to \((3m - 1)(3m + 1)\).

The power of this identity lies in its simplicity. It provides a direct method to factor many polynomial expressions without diving into more complex techniques.

Polynomial Expressions

Polynomial expressions involve multiple terms, each consisting of a variable raised to a non-negative integer power. These terms are combined using addition or subtraction. For example, \(2x^2 + 3x - 5\) is a simple polynomial expression.

Each term in a polynomial is composed of a coefficient (a number) and a variable component (like \(x^2\)). Understanding the structure of polynomials enables you to apply various algebraic techniques to simplify, expand, or evaluate them.

Polynomials can be simple, like just having two terms as in a binomial, or more complex with multiple terms—including trinomials and others. Here's a noteworthy tip: the degree of a polynomial is determined by the highest power of the variable present in the expression. For instance:

• The polynomial \(4m^2 - 9\) has a degree of 2.
• In the expression \(x^3 - 5x^2 + x\), the degree is 3.

Recognizing these characteristics helps in breaking down and analyzing the expression for further manipulation like factoring or expanding.

Algebraic Factoring

Algebraic factoring is the process of breaking down a polynomial into a product of simpler polynomials or numbers. This technique is a cornerstone in algebra, as it simplifies expressions and solves equations efficiently.

The main goal is to express the polynomial as a product of factors. To achieve this, identification of common patterns, such as the difference of squares or perfect square trinomials, comes into play. For example:

• The expression \(x^2 - 25\) can be recognized as a difference of squares and factored into \((x-5)(x+5)\).
• In other instances like \(x^2 + 5x + 6\), recognizing that it factors into \((x+2)(x+3)\) involves looking for two numbers that multiply to 6 and add up to 5.

Through factoring, seemingly complex algebraic expressions become manageable. This approach is not only used for simplifying expressions but also plays a critical role in solving quadratic equations. In essence, factoring is like unraveling the expression to see its essential parts more clearly.