Solving Algebra Word Problems with Factoring

Algebra word problems can be tricky, but they can also be a lot of fun! One of the most common types of algebra word problems involves factoring. Factoring is a powerful tool that can be used to solve a variety of problems, including those involving quadratic equations. In this blog post, we'll explore how to solve algebra word problems using factoring.

Understanding the Basics of Factoring

Factoring is the process of breaking down a polynomial into its factors. A factor is a number or expression that divides evenly into another number or expression. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12. In algebra, factoring is often used to solve quadratic equations. A quadratic equation is an equation that has a term with a variable raised to the power of 2. For example, the equation x² + 5x + 6 = 0 is a quadratic equation.

Example: The Garden Problem

Let's consider a classic algebra word problem: A rectangular garden has a length that is 3 meters longer than its width. The area of the garden is 40 square meters. What are the dimensions of the garden?

Step 1: Define the Variables

Let's define our variables:

Let w represent the width of the garden.
Let l represent the length of the garden.

Step 2: Set Up the Equations

We know that the length is 3 meters longer than the width, so we can write the equation:

l = w + 3

We also know that the area of the garden is 40 square meters. The area of a rectangle is calculated by multiplying its length and width, so we can write the equation:

l * w = 40

Step 3: Substitute and Simplify

We can substitute the first equation (l = w + 3) into the second equation to eliminate l:

(w + 3) * w = 40

Expanding the equation, we get:

w² + 3w = 40

Subtracting 40 from both sides, we get a quadratic equation:

w² + 3w - 40 = 0

Step 4: Factor the Quadratic Equation

Now, we need to factor the quadratic equation. We need to find two numbers that multiply to -40 and add up to 3. These numbers are 8 and -5. Therefore, we can factor the equation as:

(w + 8)(w - 5) = 0

Step 5: Solve for the Width

For the product of two factors to be zero, at least one of the factors must be zero. So, we have two possible solutions:

w + 8 = 0, which gives us w = -8
w - 5 = 0, which gives us w = 5

Since the width cannot be negative, we discard the first solution. Therefore, the width of the garden is 5 meters.

Step 6: Find the Length

Now that we know the width, we can use the equation l = w + 3 to find the length:

l = 5 + 3 = 8

Therefore, the length of the garden is 8 meters.

Conclusion

By factoring the quadratic equation, we were able to solve for the dimensions of the garden. This example demonstrates how factoring can be a powerful tool for solving algebra word problems. By following these steps, you can confidently tackle a variety of word problems involving quadratic equations.

Remember, practice makes perfect! The more you work with factoring, the more comfortable you'll become with solving these types of problems.

Solving Algebra Word Problems with Factoring