Factoring $81x^2 - 49$: A Difference Of Squares Guide

Let's dive into factoring the polynomial $81x^2 - 49$ using the difference of squares formula. This method is a classic and powerful tool in algebra, making complex expressions simpler to manage. In this guide, we'll break down the steps, identify the values of $a$ and $b$, and find the product of the prime factors. By the end, you'll have a solid understanding of how to apply this technique effectively.

Understanding the Difference of Squares

The difference of squares is a special pattern that emerges when you have two perfect squares separated by a subtraction sign. The general formula is:

$a^2 - b^2 = (a + b)(a - b)$

This formula tells us that if we can identify two terms that are perfect squares and are being subtracted, we can factor the expression into two binomials: one with the sum of the square roots and one with the difference of the square roots. This is incredibly useful because it turns a seemingly complex expression into a product of simpler expressions, which can make further algebraic manipulations much easier. Factoring using the difference of squares is a fundamental skill in algebra and is essential for simplifying expressions, solving equations, and understanding more advanced mathematical concepts.

Recognizing a difference of squares pattern is the first step to successful factorization. Look for terms that can be expressed as something squared. Common perfect squares include numbers like 1, 4, 9, 16, 25, and so on, as well as variables raised to even powers, such as $x^2$, $x^4$, $x^6$, and so forth. Once you've identified the pattern, applying the formula is straightforward, and the resulting factored form can be used in a variety of algebraic contexts.

Identifying $a$ and $b$ in $81x^2 - 49$

Now, let's apply this to our polynomial, $81x^2 - 49$. The first step is to recognize that both $81x^2$ and $49$ are perfect squares. We need to figure out what values, when squared, give us these terms.

For $81x^2$, we can rewrite it as $(9x)^2$. This is because $9^2 = 81$ and $(x)^2 = x^2$. So, the square root of $81x^2$ is $9x$. This means that in our formula, $a$ corresponds to $9x$.

Next, we look at $49$. We know that $7^2 = 49$, so the square root of $49$ is $7$. In our formula, $b$ corresponds to $7$.

Therefore, we have identified that $a = 9x$ and $b = 7$. Understanding how to break down the terms into their square roots is crucial for correctly applying the difference of squares formula. This process allows us to see the underlying structure of the polynomial and set up the factorization.

By identifying the values of $a$ and $b$, we are essentially reversing the squaring operation to find the base terms that, when squared, produce the terms in the original polynomial. This is a key step in factoring and allows us to rewrite the polynomial in a more manageable form. Recognizing these perfect squares and their corresponding square roots is a fundamental skill in algebra and is used extensively in various mathematical applications.

Applying the Difference of Squares Formula

Now that we know $a = 9x$ and $b = 7$, we can apply the difference of squares formula:

$a^2 - b^2 = (a + b)(a - b)$

Substituting our values for $a$ and $b$, we get:

$81x^2 - 49 = (9x + 7)(9x - 7)$

This is the factored form of the polynomial. We have successfully transformed the difference of squares into a product of two binomials. This step is the heart of the factorization process and demonstrates the power of the difference of squares formula. By recognizing the pattern and correctly identifying $a$ and $b$, we can efficiently factor complex expressions.

The factored form $(9x + 7)(9x - 7)$ represents the original polynomial in a more simplified manner. This can be particularly useful when solving equations or simplifying algebraic expressions. For instance, if we were to solve the equation $81x^2 - 49 = 0$, we could set each factor equal to zero: $9x + 7 = 0$ and $9x - 7 = 0$, and then solve for $x$. This illustrates how factoring is not just an abstract algebraic exercise but a practical tool for solving real-world problems.

Finding the Product of Prime Factors

The question also asks for the product of the prime factors. However, we need to be careful here. The factored form $(9x + 7)(9x - 7)$ involves binomial expressions, not individual prime factors. The coefficients $9$ and $7$ are part of these binomials and do not directly translate into prime factors in the traditional sense.

To clarify, the term "prime factors" typically refers to the prime numbers that divide a given integer. In the context of a polynomial, we've already factored it into its simplest form using the difference of squares formula. The binomials $(9x + 7)$ and $(9x - 7)$ are the factors, but they are not prime numbers.

If the question intended to ask for the prime factors of the coefficients within the binomials, we could break those down further. However, as the polynomial is factored completely, the focus is on the binomial factors rather than individual prime number factors.

It's important to note that the coefficients $9$ and $7$ themselves have prime factors. The prime factors of $9$ are $3$ and $3$ (since $9 = 3 imes 3$), and $7$ is already a prime number. However, these prime factors are part of the terms within the binomials and are not separate factors of the entire expression $81x^2 - 49$ in its factored form.

Conclusion

In summary, we successfully factored the polynomial $81x^2 - 49$ using the difference of squares formula. We identified $a = 9x$ and $b = 7$, and then applied the formula to get the factored form $(9x + 7)(9x - 7)$. While we discussed the prime factors of the coefficients, the main focus was on factoring the polynomial itself.

Understanding the difference of squares is crucial for simplifying algebraic expressions and solving equations. This technique provides a straightforward method for factoring expressions in the form $a^2 - b^2$, making it an indispensable tool in algebra. By recognizing the pattern and applying the formula, you can efficiently factor a wide range of polynomials.

Remember to always look for this pattern when factoring, as it can significantly simplify the process. Practice applying the difference of squares formula with various examples to build your confidence and proficiency in factoring.

For further learning and practice on factoring techniques, you might find resources on websites like Khan Academy's Algebra Section incredibly helpful. They offer a variety of lessons and exercises to reinforce your understanding of factoring and other algebraic concepts.