Problem 11 - Introductory Algebra for College Students

Question

Factor each difference of two squares. $$9-25 y^{2}$$

Step-by-Step Solution

Answer

The difference of squares \(9 - 25y^2\) can be factored into \((3 - 5y)(3 + 5y)\)

Step 1: Recognize difference of squares

The given expression is \(9 - 25y^2\). This is a difference (subtraction) of two squares because \(9 = 3^2\) and \(25y^2 = (5y)^2\). Hence this falls under the identity \(a^2 - b^2 = (a-b)(a+b)\).

Step 2: Apply the formula

The identity \(a^2 - b^2 = (a-b)(a+b)\) can be used to factorize the expression. Here, \(a = 3\) and \(b = 5y\). So applying the formula, the difference of squares becomes \((3 - 5y)(3 + 5y)\).

Key Concepts

Polynomial Factoring

Factoring polynomials is a process of breaking down a complex expression into simpler parts, much like deconstructing a building into individual blocks. It's a foundational concept in algebra which allows students to simplify expressions and solve equations more easily. For instance, if we have the polynomial expression \(9 - 25y^2\), we look for patterns or identities that enable us to transform it into a product of two or more simpler polynomials.

In our example, noticing that we're dealing with a difference of two squares can be a eureka moment, as this particular pattern is factorizable using an algebraic identity. The key to mastering polynomial factoring lies in recognizing such patterns, since once they are spotted, you can apply relevant formulas to simplify the expression rapidly. We are essentially dismantling the original expression, \(9 - 25y^2\), to reveal its core components or factors, in this case, \(3-5y\) and \(3+5y\). These factors give us a clearer, often more useful form of the original polynomial.

Algebraic Identities

Algebraic identities are like the secret codes of mathematics thrived on to unlock complex expressions with ease. They are predefined mathematical relationships that hold true for all values of their variables. One such indispensable identity is the difference of two squares, expressed as \(a^2 - b^2 = (a-b)(a+b)\).

The identity tells us that whenever you have an expression that involves the subtraction of one square from another, you can always rewrite it as the product of two binomials - one subtracting the square roots and one adding them. This tricksy identity is particularly handy because it's so common in algebra problems, including in the factoring of quadratic equations. With regular practice, the ability to recognize when this identity fits into the puzzle of a given expression becomes second nature, and as with our earlier problem, \(9 - 25y^2\), understanding this identity is essential in breaking it down to \(3 - 5y\) and \(3 + 5y\).

Quadratic Equations

Quadratic equations sit at the heart of algebra and are written in the standard form of \(ax^2 + bx + c = 0\). They commonly describe parabolic curves in coordinate geometry and arise in countless real-world applications. Factoring is one of several strategies to solve these equations, where we seek to express the quadratic expression as a product of two binomials.

In the case of a difference of two squares, such as \(9 - 25y^2\), factoring it directly provides solutions to the equation \(9 - 25y^2 = 0\). Once we factor our expression, we have \(3 - 5y\) times \(3 + 5y\) equalling zero. Employing the zero product property - which states that if a product is zero, then at least one of the factors must also be zero - leads us to two possibilities: either \(3 - 5y = 0\) or \(3 + 5y = 0\). Solving these simpler equations, we find the roots of the original quadratic equation.