Factoring Quadratic Equations: Practice Problems & Solutions

Nov 3, 2025 by Admin 61 views

Hey guys! Let's dive into the world of quadratic equations and tackle them using the factoring method. Factoring is a super useful technique for solving these equations, and we've got a bunch of practice problems to help you master it. So, grab your pencils, and let's get started!

What are Quadratic Equations and Why Factoring?

Before we jump into the problems, let's quickly recap what quadratic equations are and why factoring is such a cool method to use. A quadratic equation is basically an equation that can be written in the form ax² + bx + c = 0, where a, b, and c are constants, and a is not zero (otherwise, it wouldn't be quadratic!).

Now, why factoring? Well, factoring is a way of breaking down a quadratic expression into a product of two linear expressions. This is awesome because if we can write the equation in the form (x + p)(x + q) = 0, then we know that either (x + p) = 0 or (x + q) = 0 (or both!). This gives us the solutions to the equation super easily. It's like a magic trick, but with math!

The factoring method is a powerful tool for solving quadratic equations because it transforms a complex problem into a simpler one. By expressing the quadratic equation as a product of two binomials, we can easily identify the roots or solutions of the equation. Factoring is not only efficient but also provides a deeper understanding of the structure of quadratic equations. It's a fundamental skill in algebra that opens the door to more advanced mathematical concepts and problem-solving techniques.

For example, consider the equation x² + 5x + 6 = 0. Factoring this equation, we get (x + 2)(x + 3) = 0. Setting each factor to zero gives us x + 2 = 0 and x + 3 = 0, which lead to the solutions x = -2 and x = -3. This straightforward approach highlights the effectiveness of factoring. Guys, understanding and practicing factoring will make solving quadratic equations feel like a breeze!

Practice Problems: Let's Factor These Equations!

Okay, enough talk! Let's get our hands dirty with some actual problems. We've got 12 quadratic equations here, each with its own little twist. We'll break them down step-by-step so you can see how the factoring method works in action.

1) x² = 64

Step 1: Rewrite the equation in standard form. To use factoring, we need the equation in the form ax² + bx + c = 0. So, let's subtract 64 from both sides: x² - 64 = 0.
Step 2: Recognize the difference of squares. This looks like a classic difference of squares pattern: a² - b² = (a + b)(a - b). In our case, a = x and b = 8.
Step 3: Factor the equation. Using the pattern, we get (x + 8)(x - 8) = 0.
Step 4: Set each factor to zero and solve. x + 8 = 0 => x = -8 x - 8 = 0 => x = 8
Solution: x = -8, 8

2) x² - 100 = 0

Step 1: Recognize the difference of squares. Again, we've got a difference of squares! Here, a = x and b = 10.
Step 2: Factor the equation. This gives us (x + 10)(x - 10) = 0.
Step 3: Set each factor to zero and solve. x + 10 = 0 => x = -10 x - 10 = 0 => x = 10
Solution: x = -10, 10

3) x² - 9x = 0

Step 1: Factor out the common factor. In this case, both terms have an x, so we can factor that out: x(x - 9) = 0.
Step 2: Set each factor to zero and solve. x = 0 x - 9 = 0 => x = 9
Solution: x = 0, 9

4) 4x² - 25 = 0

Step 1: Recognize the difference of squares. This is another difference of squares! Here, a = 2x and b = 5.
Step 2: Factor the equation. We get (2x + 5)(2x - 5) = 0.
Step 3: Set each factor to zero and solve. 2x + 5 = 0 => 2x = -5 => x = -5/2 2x - 5 = 0 => 2x = 5 => x = 5/2
Solution: x = -5/2, 5/2

5) x² - 4x + 3 = 0

Step 1: Find two numbers that multiply to c (3) and add up to b (-4). The numbers -1 and -3 fit the bill (-1 * -3 = 3 and -1 + -3 = -4).
Step 2: Factor the equation. We can write this as (x - 1)(x - 3) = 0.
Step 3: Set each factor to zero and solve. x - 1 = 0 => x = 1 x - 3 = 0 => x = 3
Solution: x = 1, 3

6) x² + 2x - 3 = 0

Step 1: Find two numbers that multiply to c (-3) and add up to b (2). The numbers 3 and -1 work (3 * -1 = -3 and 3 + -1 = 2).
Step 2: Factor the equation. This factors to (x + 3)(x - 1) = 0.
Step 3: Set each factor to zero and solve. x + 3 = 0 => x = -3 x - 1 = 0 => x = 1
Solution: x = -3, 1

7) x² + 3x - 10 = 0

Step 1: Find two numbers that multiply to c (-10) and add up to b (3). The numbers 5 and -2 satisfy these conditions (5 * -2 = -10 and 5 + -2 = 3).
Step 2: Factor the equation. We can factor the equation as (x + 5)(x - 2) = 0.
Step 3: Set each factor to zero and solve. x + 5 = 0 => x = -5 x - 2 = 0 => x = 2
Solution: x = -5, 2

8) x² - 6x + 5 = 0

Step 1: Find two numbers that multiply to c (5) and add up to b (-6). The numbers -1 and -5 work perfectly (-1 * -5 = 5 and -1 + -5 = -6).
Step 2: Factor the equation. This gives us (x - 1)(x - 5) = 0.
Step 3: Set each factor to zero and solve. x - 1 = 0 => x = 1 x - 5 = 0 => x = 5
Solution: x = 1, 5

9) x² - 3x + 2 = 0

Step 1: Find two numbers that multiply to c (2) and add up to b (-3). The numbers -1 and -2 fit the criteria (-1 * -2 = 2 and -1 + -2 = -3).
Step 2: Factor the equation. The factored form is (x - 1)(x - 2) = 0.
Step 3: Set each factor to zero and solve. x - 1 = 0 => x = 1 x - 2 = 0 => x = 2
Solution: x = 1, 2

10) -x² + 6x = 0

Step 1: Factor out the common factor and a -1. We can factor out a -x, which gives us -x(x - 6) = 0.
Step 2: Set each factor to zero and solve. -x = 0 => x = 0 x - 6 = 0 => x = 6
Solution: x = 0, 6

11) x² + 18x = -81

Step 1: Rewrite the equation in standard form. Add 81 to both sides to get x² + 18x + 81 = 0.
Step 2: Recognize the perfect square trinomial. This equation is a perfect square trinomial: (x + 9)² = 0.
Step 3: Solve for x. x + 9 = 0 => x = -9
Solution: x = -9 (This is a repeated root).

12) x² + 3x = 18

Step 1: Rewrite the equation in standard form. Subtract 18 from both sides: x² + 3x - 18 = 0.
Step 2: Find two numbers that multiply to c (-18) and add up to b (3). The numbers 6 and -3 work (6 * -3 = -18 and 6 + -3 = 3).
Step 3: Factor the equation. This factors to (x + 6)(x - 3) = 0.
Step 4: Set each factor to zero and solve. x + 6 = 0 => x = -6 x - 3 = 0 => x = 3
Solution: x = -6, 3

Key Takeaways for Solving Quadratic Equations by Factoring

Let's wrap up with the key things we've learned from tackling these problems:

Standard Form is Your Friend: Always rewrite the equation in the form ax² + bx + c = 0. This sets you up perfectly for factoring.
Spotting Patterns is Crucial: Keep an eye out for special patterns like the difference of squares (a² - b² = (a + b)(a - b)) and perfect square trinomials. Recognizing these patterns makes factoring way easier.
Find Those Numbers: When factoring trinomials, finding the two numbers that multiply to c and add to b is the name of the game. Practice this, and you'll become a factoring whiz in no time!
Don't Forget the Common Factor: Always check if you can factor out a common factor first. This simplifies the equation and makes the rest of the factoring process smoother.
Set Factors to Zero: Once you've factored the equation, remember the magic step: set each factor equal to zero. This is what gives you the solutions!

Quadratic equations might seem intimidating at first, but with practice, you'll become a factoring master! Guys, keep practicing, and you'll be solving these equations in your sleep. You got this! Factoring quadratic equations becomes much easier with consistent practice. The more you practice, the quicker you'll identify the patterns and the numbers needed for factoring. Remember, every mistake is a learning opportunity, so don't be discouraged by challenges. Keep pushing, and you'll develop a strong intuition for factoring, making it an invaluable skill in your mathematical journey. So, keep up the fantastic work, and happy factoring!

I hope this helps you understand factoring quadratic equations better. Keep practicing, and you'll become a pro in no time! Let me know if you have any more questions.