Factorizar A²x² + (a² - B²)x - B²: Guía Paso A Paso

¡Hola, math whizzes! Today, we're diving deep into the awesome world of algebra to tackle a factorization problem that might look a bit intimidating at first glance: factorizar a²x² + (a² - b²)x - b². Don't sweat it, guys! By the end of this article, you'll be factoring this bad boy like a pro. We'll break it down step-by-step, so even if you're just getting your feet wet with quadratic expressions, you'll be able to follow along and, more importantly, understand what's going on. Ready to flex those algebraic muscles? Let's get started!

Understanding the Expression: What Are We Dealing With?

So, first things first, let's get acquainted with the expression we need to factor: a²x² + (a² - b²)x - b². This is a quadratic expression in terms of 'x', meaning the highest power of 'x' is 2. The coefficients involve 'a' and 'b', which are constants in this context. When we talk about factorizing an expression, we're essentially trying to break it down into a product of simpler expressions, usually binomials. Think of it like finding the building blocks of a mathematical structure. For this particular expression, we're looking for two binomials that, when multiplied together, give us exactly a²x² + (a² - b²)x - b². The presence of $a^2$ and $b^2$ hints that we might be dealing with differences of squares or something related to perfect squares. This is a common pattern in algebra, and recognizing these patterns is a superpower when it comes to factorization. We'll explore different methods to peel back the layers of this expression and reveal its fundamental factors. Remember, algebra is like a puzzle, and factorization is one of the most satisfying ways to solve it!

Method 1: The 'AC' Method (or Grouping)

The 'AC' method is a classic technique for factoring quadratic trinomials of the form $Ax^2 + Bx + C$. In our case, $A = a^2$, $B = (a^2 - b^2)$, and $C = -b^2$. The goal is to find two numbers that multiply to $A imes C$ and add up to $B$. So, we need to find two numbers that multiply to $(a^2) imes (-b^2) = -a^2b^2$ and add up to $(a^2 - b^2)$.

Let's think about the factors of $-a^2b^2$. We can have pairs like $(a^2, -b^2)$, $(-a^2, b^2)$, $(ab, -ab)$, etc. We're looking for a pair that adds up to $a^2 - b^2$. If we consider the pair $(a^2, -b^2)$, their sum is $a^2 + (-b^2) = a^2 - b^2$. Bingo! We found our two numbers: $a^2$ and $-b^2$.

Now, we rewrite the middle term, $(a^2 - b^2)x$, using these two numbers: $a^2x - b^2x$. So, our expression becomes:

$a^2x^2 + a^2x - b^2x - b^2$

Next, we group the terms in pairs and factor out the greatest common factor (GCF) from each pair.

From the first pair, $a^2x^2 + a^2x$, the GCF is $a^2x$. Factoring it out, we get: $a^2x(x + 1)$.

From the second pair, $-b^2x - b^2$, the GCF is $-b^2$. Factoring it out, we get: $-b^2(x + 1)$.

Now our expression looks like this:

$a^2x(x + 1) - b^2(x + 1)$

Notice that we have a common binomial factor of $(x + 1)$. We can now factor this out:

$(x + 1)(a^2x - b^2)$

And there you have it! We've successfully factored the expression using the 'AC' method. This method is super reliable for quadratics. The key is to correctly identify the two numbers that satisfy the product and sum conditions.

Method 2: Recognizing Patterns (Difference of Squares Hint)

Sometimes, you can spot patterns that lead to a quicker factorization. Let's look at our expression again: a²x² + (a² - b²)x - b². The terms $a^2x^2$ and $-b^2$ look like perfect squares. What if we try to arrange it in a way that resembles a difference of squares or a perfect square trinomial?

Let's rewrite the middle term slightly differently. We know that $(a^2 - b^2)x = a^2x - b^2x$. Substituting this back into the original expression:

$a^2x^2 + a^2x - b^2x - b^2$

Now, let's try grouping differently. If we group the first and second terms, and the third and fourth terms:

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

Factor out the GCF from each group:

$a^2x(x + 1) - b^2(x + 1)$

As we saw in the previous method, this leads us to $(x + 1)(a^2x - b^2)$.

What if we tried to think about it in terms of $(ax)^2$ and $b^2$? Our expression is $a^2x^2 + (a^2 - b^2)x - b^2$. This doesn't immediately scream 'difference of squares' in its current form. However, the presence of $a^2$ and $b^2$ as separate terms suggests we might be able to factor out something involving 'a' and 'b' from parts of the expression.

Let's consider the structure again. We have a term with $x^2$, a term with $x$, and a constant term. This is the standard form of a quadratic. The coefficients are where the magic happens. The coefficient of $x$ is $(a^2 - b^2)$. This difference of squares pattern in the coefficient itself is a huge clue. It suggests that perhaps the factors will involve $a$ and $b$ in a way that reflects this difference.

Think about a generic quadratic $(px + q)(rx + s)$. Expanding this gives $prx^2 + (ps + qr)x + qs$. We need to match this to $a^2x^2 + (a^2 - b^2)x - b^2$.

So, we need:

• $pr = a^2$
• $qs = -b^2$
• $ps + qr = a^2 - b^2$

Let's try some possibilities for $p, q, r, s$ that fit the first two conditions and see if they satisfy the third.

If we let $p=a$ and $r=a$, then $pr=a^2$. This works.

For $qs=-b^2$, we could have $q=a$ and $s=-b$, or $q=-a$ and $s=b$, or $q=b$ and $s=-a$, or $q=-b$ and $s=a$. This is getting complicated.

Let's reconsider our factors $(x + 1)(a^2x - b^2)$. If we expand this:

$(x)(a^2x) + (x)(-b^2) + (1)(a^2x) + (1)(-b^2)$

$= a^2x^2 - b^2x + a^2x - b^2$

Rearranging the middle terms:

$= a^2x^2 + a^2x - b^2x - b^2$

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

This matches our original expression! So, the pattern recognition here wasn't about seeing a direct difference of squares for the entire expression, but rather recognizing that the components ($a^2x^2$, $-b^2$) and the structure of the middle term's coefficient $(a^2 - b^2)$ strongly suggested factors involving $a^2$ and $b^2$ in specific ways. It requires a bit of intuition and practice to see how these pieces fit together. Sometimes, a seemingly complex coefficient is the key clue.

Method 3: Using the Quadratic Formula (as a check or alternative)

While factorization is usually about finding factors directly, the quadratic formula can sometimes help if you're stuck or want to verify your answer. For a quadratic equation $Ax^2 + Bx + C = 0$, the solutions for $x$ are given by:

$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

If we set our expression equal to zero: $a^2x^2 + (a^2 - b^2)x - b^2 = 0$. Here, $A = a^2$, $B = (a^2 - b^2)$, and $C = -b^2$.

Let's plug these into the quadratic formula:

$x = \frac{-(a^2 - b^2) \pm \sqrt{(a^2 - b^2)^2 - 4(a^2)(-b^2)}}{2(a^2)}$

$x = \frac{-a^2 + b^2 \pm \sqrt{(a^4 - 2a^2b^2 + b^4) + 4a^2b^2}}{2a^2}$

$x = \frac{-a^2 + b^2 \pm \sqrt{a^4 + 2a^2b^2 + b^4}}{2a^2}$

The term under the square root, $a^4 + 2a^2b^2 + b^4$, is a perfect square trinomial! It's $(a^2 + b^2)^2$.

So, we have:

$x = \frac{-a^2 + b^2 \pm \sqrt{(a^2 + b^2)^2}}{2a^2}$

$x = \frac{-a^2 + b^2 \pm (a^2 + b^2)}{2a^2}$

Now we have two possible values for $x$:

Case 1: Using the '+' sign $x_1 = \frac{-a^2 + b^2 + (a^2 + b^2)}{2a^2} = \frac{-a^2 + b^2 + a^2 + b^2}{2a^2} = \frac{2b^2}{2a^2} = \frac{b^2}{a^2}$

Case 2: Using the '-' sign $x_2 = \frac{-a^2 + b^2 - (a^2 + b^2)}{2a^2} = \frac{-a^2 + b^2 - a^2 - b^2}{2a^2} = \frac{-2a^2}{2a^2} = -1$

So, the roots of the equation $a^2x^2 + (a^2 - b^2)x - b^2 = 0$ are $x = \frac{b^2}{a^2}$ and $x = -1$.

If the roots of a quadratic $Ax^2 + Bx + C = 0$ are $x_1$ and $x_2$, then the factored form is $A(x - x_1)(x - x_2)$. In our case, $A = a^2$, $x_1 = \frac{b^2}{a^2}$, and $x_2 = -1$.

So, the factored form is:

$a^2(x - \frac{b^2}{a^2})(x - (-1))$

$a^2(x - \frac{b^2}{a^2})(x + 1)$

Now, let's distribute the $a^2$ into the first factor to get rid of the fraction:

$(a^2x - a^2\frac{b^2}{a^2})(x + 1)$

$(a^2x - b^2)(x + 1)$

This gives us the same result: $(x + 1)(a^2x - b^2)$. The quadratic formula is a powerful tool, especially when direct factorization methods seem tricky. It confirms our previous findings and shows the interconnectedness of roots and factors.

Final Answer and Why It Works

So, after exploring a few different paths, we've arrived at the same, beautiful factorization: (x + 1)(a²x - b²). This is the simplified form of a²x² + (a² - b²)x - b².

Why does this work? Remember, when you multiply these two binomials together, you should get back the original trinomial. Let's quickly check:

$(x + 1)(a^2x - b^2)$

Using the FOIL method (First, Outer, Inner, Last):

• First: $x imes a^2x = a^2x^2$
• Outer: $x imes (-b^2) = -b^2x$
• Inner: $1 imes a^2x = a^2x$
• Last: $1 imes (-b^2) = -b^2$

Now, add all these terms together:

$a^2x^2 - b^2x + a^2x - b^2$

Combine the like terms (the terms with 'x'):

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

Boom! We're back to our original expression. This confirms that our factorization is absolutely correct. It's like taking apart a LEGO structure and then putting it back together to make sure all the pieces fit perfectly. Each method we used – the 'AC' method, pattern recognition, and the quadratic formula – led us to this exact result, reinforcing its accuracy. Math is all about finding these consistent truths!

Conclusion

Factorizing expressions like a²x² + (a² - b²)x - b² might seem daunting initially, but with the right techniques and a little practice, it becomes much more manageable. We've seen how the 'AC' method, careful observation of patterns, and even the quadratic formula can lead you to the solution. The key is to stay organized, understand the properties of algebraic expressions, and not be afraid to try different approaches. Keep practicing, guys, and you'll find that algebraic factorization is not just a skill, but a really cool way to understand the structure of mathematical expressions. Happy factoring!