Problem 65 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$9 y^{2}-64$$

Step-by-Step Solution

Answer

The factorization of the polynomial \(9y^{2}-64\) is \((3y - 8)(3y + 8)\).

Step 1: Identify the Difference of Squares

The given polynomial \(9y^{2}-64\) is a difference of two squares. In this case, \(a\) is \(3y\) and \(b\) is \(8\), because \((3y)^2 = 9y^2\) and \(8^2 = 64\).

Step 2: Apply the Difference of Squares Formula

Substitute \(3y\) for \(a\) and \(8\) for \(b\) in the formula \(a^2 - b^2 = (a - b)(a + b)\). This results in the factored form: \((3y - 8)(3y + 8)\).

Step 3: Verify the Factorization

To verify, multiply the binomials \((3y - 8)\) and \((3y + 8)\) to confirm the original polynomial is obtained. \((3y - 8)(3y + 8) = 9y^{2}-64\). This confirms the factorization is correct.

Key Concepts

Difference of Squares

The difference of squares is an important property in algebra where two squared terms are subtracted from one another, and it has a unique factorization pattern. For any two terms, with variables or numbers, if one term is the square of something and the second term is also the square of something else, you can rewrite the expression as
\((a^2 - b^2) = (a - b)(a + b)\).

In our example, we see that \(9y^{2}\) is the square of \(3y\) and \(64\) is the square of \(8\), so we can apply this pattern. Recognizing this pattern simplifies the process, allowing us to quickly factor expressions that might initially appear complicated. This is a fundamental concept in polynomial factorization that students often use in algebra.

Polynomial Factorization

Polynomial factorization, simply put, is the process of breaking down a complex polynomial into a product of simpler polynomials that, when multiplied together, give you the original polynomial. Think of it as a reverse multiplication process specific to polynomials. In our example, \(9y^{2}-64\) was factored into \((3y - 8)(3y + 8)\), indicating that these two simpler binomials multiply to give the original expression.

It's like taking a lego structure and finding the individual blocks that made it up. There are several factoring techniques including, but not limited to, extracting common factors, difference of squares, completing the square, and trinomial factoring. Understanding when and how to apply these techniques can greatly streamline solving and understanding algebraic problems.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can consist of numbers, variables, and operations (like addition and subtraction). These expressions can become quite elaborate and often need to be simplified or manipulated to solve equations. In our scenario, the algebraic expression is \(9y^{2}-64\), a difference of squares.

In order to work with these expressions effectively, students must be familiar with various laws and properties of algebra, such as the distributive property, associative property, and commutative property. Being comfortable with these concepts is crucial for success in solving algebraic problems. Making sense of these expressions, and being adept at manipulating them, is one of the core skills in algebra.

Verifying Factorizations

Verifying factorizations is just as important as the factorization process itself. It's a step that confirms whether or not the factorization is correct. You can think of it as checking your work after solving a puzzle; ensuring all pieces are in the right place.

One straightforward way to verify the factorization of a polynomial is by performing the multiplication of the factors. If you end up with the original polynomial, you've done it right! In the context of the exercise, multiplying the factors \((3y - 8)\) and \((3y + 8)\) gets us back to the original expression \(9y^{2}-64\), confirming that our factorization is indeed correct. This practice is an excellent way to avoid errors and ensure accuracy when dealing with algebraic expressions.