Factoring Quadratics: Step-by-Step Guide With Examples

by Henrik Larsen 55 views

Hey guys! Factoring quadratic expressions might seem daunting at first, but with a systematic approach and a little practice, you'll be solving these problems like a pro. In this comprehensive guide, we'll break down the process step-by-step, using several examples to illustrate the techniques. We'll tackle expressions of the form ax² + bx + c, where a, b, and c are constants. Let's dive in!

Understanding Quadratic Expressions

Before we jump into factoring, let's make sure we're all on the same page about what a quadratic expression is. A quadratic expression is a polynomial expression of degree two. This means the highest power of the variable (usually 'x') is 2. The general form of a quadratic expression is ax² + bx + c, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. For example, 6x² + 19x + 15, 6x² - 19x + 15, and 6x² + x - 15 are all quadratic expressions.

The goal of factoring a quadratic expression is to rewrite it as a product of two linear expressions. A linear expression is a polynomial expression of degree one, meaning the highest power of the variable is 1. For instance, (3x + 5) and (2x + 3) are linear expressions. When we factor a quadratic expression, we're essentially reversing the process of expansion (or multiplication). Think of it like this: if we multiply (3x + 5) and (2x + 3), we get 6x² + 19x + 15. Factoring is the process of going from 6x² + 19x + 15 back to (3x + 5)(2x + 3).

Factoring quadratic expressions is a fundamental skill in algebra and has numerous applications in solving equations, simplifying expressions, and graphing functions. It's a crucial stepping stone for more advanced mathematical concepts, so mastering it is well worth the effort. Now, let's get into the nitty-gritty of factoring techniques!

Method 1: Factoring by Grouping (The AC Method)

The most common and versatile method for factoring quadratic expressions is called factoring by grouping, also known as the AC method. This method works for a wide range of quadratic expressions, including those where the leading coefficient (the 'a' value) is not equal to 1. Here's the breakdown of how it works:

  1. Identify a, b, and c: First, identify the coefficients a, b, and c in the quadratic expression ax² + bx + c. For example, in the expression 6x² + 19x + 15, a = 6, b = 19, and c = 15.
  2. Calculate AC: Multiply the coefficients 'a' and 'c'. In our example, AC = 6 * 15 = 90.
  3. Find Two Factors: Find two factors of AC (90 in our case) that add up to 'b' (which is 19). This is the crucial step! You might need to list out the factor pairs of AC to find the right combination. For 90, the factor pairs are (1, 90), (2, 45), (3, 30), (5, 18), (6, 15), and (9, 10). The pair (9, 10) works because 9 + 10 = 19.
  4. Rewrite the Middle Term: Rewrite the middle term (bx) using the two factors you just found. In our example, we rewrite 19x as 9x + 10x. So, the expression becomes 6x² + 9x + 10x + 15.
  5. Factor by Grouping: Now, group the first two terms and the last two terms together: (6x² + 9x) + (10x + 15). Factor out the greatest common factor (GCF) from each group. From the first group, the GCF is 3x, so we get 3x(2x + 3). From the second group, the GCF is 5, so we get 5(2x + 3). The expression now looks like this: 3x(2x + 3) + 5(2x + 3).
  6. Final Factorization: Notice that both terms now have a common factor of (2x + 3). Factor this out: (2x + 3)(3x + 5). And there you have it! The factored form of 6x² + 19x + 15 is (2x + 3)(3x + 5).

This method might seem like a lot of steps, but with practice, it becomes second nature. Let's apply this method to the examples you provided.

Applying the AC Method to Your Examples

Let's break down each of the examples you provided, step-by-step, using the AC method. This will give you a clear understanding of how to apply the technique in different scenarios.

(1) 6x² + 19x + 15

We already walked through this example in detail above, but let's recap:

  • a = 6, b = 19, c = 15
  • AC = 6 * 15 = 90
  • Factors of 90 that add up to 19: 9 and 10
  • Rewrite the middle term: 6x² + 9x + 10x + 15
  • Factor by grouping: 3x(2x + 3) + 5(2x + 3)
  • Final factorization: (2x + 3)(3x + 5)

So, 6x² + 19x + 15 = (3x + 5)(2x + 3), which matches the answer you provided!

(2) 6x² - 19x + 15

Notice the change in sign in the middle term. This will affect the factors we need to find.

  • a = 6, b = -19, c = 15
  • AC = 6 * 15 = 90
  • Factors of 90 that add up to -19: -9 and -10 (Remember, two negative numbers multiply to a positive number)
  • Rewrite the middle term: 6x² - 9x - 10x + 15
  • Factor by grouping: 3x(2x - 3) - 5(2x - 3)
  • Final factorization: (2x - 3)(3x - 5)

Therefore, 6x² - 19x + 15 = (3x - 5)(2x - 3), which also matches your answer.

(3) 6x² + x - 15

Now, let's tackle one where the constant term is negative. This introduces another slight twist to the process.

  • a = 6, b = 1, c = -15
  • AC = 6 * (-15) = -90
  • Factors of -90 that add up to 1: 10 and -9 (One factor must be positive, and the other negative)
  • Rewrite the middle term: 6x² + 10x - 9x - 15
  • Factor by grouping: 2x(3x + 5) - 3(3x + 5)
  • Final factorization: (3x + 5)(2x - 3)

So, 6x² + x - 15 = (3x + 5)(2x - 3).

(4) 6x² - x - 15

This is very similar to the previous example, but the sign of the middle term is different. This will change which factors we use.

  • a = 6, b = -1, c = -15
  • AC = 6 * (-15) = -90
  • Factors of -90 that add up to -1: -10 and 9
  • Rewrite the middle term: 6x² - 10x + 9x - 15
  • Factor by grouping: 2x(3x - 5) + 3(3x - 5)
  • Final factorization: (3x - 5)(2x + 3)

Thus, 6x² - x - 15 = (3x - 5)(2x + 3).

(5) 6x² + 5x - 21

Let's keep practicing! This one also has a negative constant term.

  • a = 6, b = 5, c = -21
  • AC = 6 * (-21) = -126
  • Factors of -126 that add up to 5: 14 and -9
  • Rewrite the middle term: 6x² + 14x - 9x - 21
  • Factor by grouping: 2x(3x + 7) - 3(3x + 7)
  • Final factorization: (3x + 7)(2x - 3)

Therefore, 6x² + 5x - 21 = (3x + 7)(2x - 3).

(6) 6x² - 5x - 21

And finally, let's tackle the last one. Again, the change in the sign of the middle term is key.

  • a = 6, b = -5, c = -21
  • AC = 6 * (-21) = -126
  • Factors of -126 that add up to -5: -14 and 9
  • Rewrite the middle term: 6x² - 14x + 9x - 21
  • Factor by grouping: 2x(3x - 7) + 3(3x - 7)
  • Final factorization: (3x - 7)(2x + 3)

So, 6x² - 5x - 21 = (3x - 7)(2x + 3).

Tips and Tricks for Factoring

Factoring can be tricky, but here are some tips and tricks to make the process smoother:

  • Always look for a Greatest Common Factor (GCF) first: Before applying any other factoring method, check if there's a GCF that can be factored out from all the terms. This simplifies the expression and makes it easier to factor further. For example, if you have 12x² + 18x + 6, you can factor out a 6 first, resulting in 6(2x² + 3x + 1).
  • Pay attention to the signs: The signs of the coefficients (a, b, and c) can give you clues about the signs of the factors. If 'c' is positive, both factors will have the same sign (either both positive or both negative). If 'c' is negative, the factors will have opposite signs. The sign of 'b' tells you which sign will have a larger magnitude.
  • Practice, practice, practice: The more you practice factoring, the better you'll become at recognizing patterns and applying the appropriate techniques. Work through plenty of examples, and don't be afraid to make mistakes – that's how you learn!
  • Check your answer: After factoring, you can always check your answer by multiplying the factors back together. If you get the original quadratic expression, you've factored it correctly.

Conclusion

Factoring quadratic expressions is a fundamental algebraic skill that unlocks a world of problem-solving possibilities. By mastering the AC method and practicing regularly, you'll gain confidence in your ability to tackle these problems. Remember to break down each problem step-by-step, pay attention to the signs, and don't hesitate to double-check your work. Keep practicing, and you'll become a factoring whiz in no time!