Factor Completely: 81x² - 49

Factoring is a fundamental skill in algebra, allowing us to break down complex expressions into simpler, multiplied components. This process is akin to deconstructing a number into its prime factors, like breaking down 12 into 2 x 2 x 3. When we're asked to factor completely, it means we need to go as far as possible in this decomposition, ensuring each factor cannot be factored further.

Today, we're going to tackle a specific expression: 81x² - 49. This might look a little intimidating at first glance, with its coefficients and the variable squared, but it's a perfect example of a common factoring pattern that, once recognized, makes the problem surprisingly straightforward. We'll explore the method, understand why it works, and ensure we achieve a complete factorization.

Recognizing the Difference of Squares

The key to factoring 81x² - 49 lies in recognizing a specific algebraic pattern: the difference of squares. This pattern appears when you have two terms, both of which are perfect squares, separated by a subtraction sign. The general form of a difference of squares is $a^2 - b^2$. If you encounter an expression in this form, you can always factor it into $(a - b)(a + b)$.

Let's break down why this formula works. If we were to expand $(a - b)(a + b)$ using the distributive property (often remembered by the FOIL method for binomials), we'd get:

$(a - b)(a + b) = a(a + b) - b(a + b)$ $= a imes a + a imes b - b imes a - b imes b$ $= a^2 + ab - ba - b^2$

Since multiplication is commutative ($ab$ is the same as $ba$), the middle terms cancel each other out: $ab - ba = 0$. This leaves us with $a^2 - b^2$, confirming that our factorization is correct.

Now, let's apply this to our expression, 81x² - 49. We need to identify what 'a' and 'b' are in this context. For the expression to fit the $a^2 - b^2$ pattern, both 81x² and 49 must be perfect squares.

• Is 81x² a perfect square? Yes, it is. The square root of 81 is 9, and the square root of $x^2$ is $x$. So, $(9x)^2 = 81x^2$. This means our 'a' term is $9x$.
• Is 49 a perfect square? Yes, it is. The square root of 49 is 7. So, $(7)^2 = 49$. This means our 'b' term is $7$.

Since we have confirmed that 81x² is $(9x)^2$ and 49 is $(7)^2$, and they are separated by a subtraction sign, our expression 81x² - 49 perfectly matches the difference of squares pattern $a^2 - b^2$, where $a = 9x$ and $b = 7$.

Therefore, we can apply the formula $(a - b)(a + b)$ directly. Substituting $9x$ for 'a' and $7$ for 'b', we get:

$(9x - 7)(9x + 7)$

This is the completely factored form of 81x² - 49 because neither $(9x - 7)$ nor $(9x + 7)$ can be factored any further using simple algebraic techniques. They are binomials with no common factors, and they don't fit any other standard factoring patterns.

Ensuring Complete Factorization

When asked to factor completely, it's crucial to double-check that each of the resulting factors cannot be broken down any further. This means examining each factor for common factors, or whether it itself can be factored using other methods like the difference of squares, sum/difference of cubes, or trinomial factoring.

In our case, we arrived at the factors $(9x - 7)$ and $(9x + 7)$. Let's analyze them:

• Factor 1: $(9x - 7)$

  • Common Factors: Are there any numbers or variables that divide both $9x$ and $7$? The factors of 9 are 1, 3, 9. The factors of 7 are 1, 7. The greatest common factor (GCF) is just 1. There's no common variable factor. So, no common factors other than 1.
  • Other Patterns: Does $(9x - 7)$ fit any other factoring patterns? It's a binomial, but it's not a difference of squares (because $9x$ is a square, but 7 is not a perfect square of a simple term like $k^2$). It's not a sum or difference of cubes. Therefore, $(9x - 7)$ is already in its simplest factored form.

• Factor 2: $(9x + 7)$

  • Common Factors: Similar to the first factor, the GCF of $9x$ and $7$ is 1. There are no common variable factors.
  • Other Patterns: This is a binomial, but it's not a difference of squares (because of the plus sign, it's a sum). It's not a sum or difference of cubes. Therefore, $(9x + 7)$ is also in its simplest factored form.

Since both factors, $(9x - 7)$ and $(9x + 7)$, are irreducible (cannot be factored further), we can be confident that our factorization of $81x^2 - 49$ into $(9x - 7)(9x + 7)$ is indeed complete.

This process highlights the importance of recognizing patterns in algebra. The difference of squares pattern is one of the most fundamental and frequently used, so familiarizing yourself with it will save you a lot of time and effort when factoring.

Steps to Factor Completely 81x² - 49

Let's summarize the steps we took to ensure we factored 81x² - 49 completely:

1. Identify the Expression: We started with the expression 81x² - 49.
2. Look for Patterns: We scanned the expression for common factoring patterns. The presence of two terms, both being perfect squares, separated by a minus sign immediately suggested the difference of squares pattern ($a^2 - b^2$).
3. Determine 'a' and 'b':
   • We identified the first term, 81x², as a perfect square: $(9x)^2$. So, $a = 9x$.
   • We identified the second term, 49, as a perfect square: $(7)^2$. So, $b = 7$.
4. Apply the Formula: We used the difference of squares factorization formula: $a^2 - b^2 = (a - b)(a + b)$. Substituting our values for 'a' and 'b', we got $(9x - 7)(9x + 7)$.
5. Check for Completeness: We examined each factor, $(9x - 7)$ and $(9x + 7)$, to see if they could be factored further. We found no common factors (other than 1) and confirmed they did not fit any other standard factoring patterns. Thus, the factorization is complete.

Following these steps systematically helps ensure that you don't miss any opportunities for factorization and arrive at the correct, complete answer.

Alternative Scenario: What if there were a Greatest Common Factor (GCF)?

It's important to remember that before applying specific patterns like the difference of squares, you should always check for a Greatest Common Factor (GCF) among all the terms in the expression. If a GCF exists, you factor it out first. The remaining expression might then be factorable using other methods.

For our expression, 81x² - 49, let's consider the coefficients (81 and 49) and the variable terms ($x^2$ and none). The factors of 81 are 1, 3, 9, 27, 81. The factors of 49 are 1, 7, 49. The only common factor between 81 and 49 is 1. Since there's no common numerical factor greater than 1, and the second term doesn't have an 'x', the GCF of the entire expression 81x² - 49 is simply 1.

Because the GCF is 1, factoring it out doesn't change the expression. This confirms that our next step should indeed be to look for other factoring patterns, which led us to the difference of squares.

Imagine, for a moment, if the expression was 16x² - 100. In this case:

• The GCF of 16 and 100 is 4.
• So, we would first factor out 4: $4(4x^2 - 25)$.
• Now, we look at the expression inside the parentheses: $4x^2 - 25$. This is a difference of squares!
  • $a^2 = 4x^2 ightarrow a = 2x$
  • $b^2 = 25 ightarrow b = 5$
• So, $4x^2 - 25$ factors into $(2x - 5)(2x + 5)$.
• The complete factorization of 16x² - 100 would then be $4(2x - 5)(2x + 5)$.

This example demonstrates why checking for the GCF first is a crucial step in ensuring complete factorization. In the case of 81x² - 49, the GCF was just 1, so the difference of squares was applied directly to the original expression.

Conclusion

Factoring the expression 81x² - 49 completely relies on recognizing and applying the difference of squares pattern. By identifying $81x^2$ as $(9x)^2$ and $49$ as $(7)^2$, we can use the formula $a^2 - b^2 = (a - b)(a + b)$ to arrive at the factored form $(9x - 7)(9x + 7)$. We have verified that both resulting binomials are irreducible, confirming that the factorization is complete. This technique is a cornerstone of algebraic manipulation, essential for solving equations and simplifying expressions. For more practice with algebraic factoring, resources like Khan Academy offer excellent tutorials and exercises. You can also find comprehensive guides on factoring techniques at Math is Fun.