Factoring Polynomials: X^4 Y - 4x^2 Y - 5y

by Andrew McMorgan 43 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of algebra to tackle a factoring problem that might look a little intimidating at first glance, but trust me, it's totally doable. We're going to break down how to find the completely factored form of $x^4 y - 4x^2 y - 5y$. This isn't just about getting the right answer; it's about understanding the process and building your confidence with polynomial expressions. So, grab your notebooks, maybe a snack, and let's get this party started!

Step 1: Finding the Greatest Common Factor (GCF)

The first rule of factoring, my friends, is always to look for the greatest common factor (GCF). This is the biggest 'chunk' that can be divided out from every term in the expression. Let's look at our expression: $x^4 y - 4x^2 y - 5y$. We've got three terms here: $x^4 y$, $-4x^2 y$, and $-5y$. Now, let's inspect the coefficients first: 1, -4, and -5. The only number that divides evenly into all of these is 1. So, no numerical GCF other than 1. But what about the variables? We have $x^4$, $x^2$, and $y$ in the first term, $x^2$ and $y$ in the second, and just $y$ in the third. Aha! We see that 'y' is present in every single term. Therefore, 'y' is our GCF. We're going to factor this 'y' out right at the beginning. When we divide each term by 'y', we get:

x4yy=x4 \frac{x^4 y}{y} = x^4

−4x2yy=−4x2 \frac{-4x^2 y}{y} = -4x^2

−5yy=−5 \frac{-5y}{y} = -5

So, after factoring out the GCF, our expression transforms into $y(x^4 - 4x^2 - 5)$. This is a huge step because it simplifies the problem considerably. Always, always start by pulling out that GCF, guys. It makes the rest of the factoring journey so much smoother. Now, our mission is to factor the expression inside the parentheses: $x^4 - 4x^2 - 5$. This is where the real fun begins, and we'll explore different techniques to conquer this part in the next section.

Step 2: Factoring the Quadratic Form

Alright, so we've got our expression down to $y(x^4 - 4x^2 - 5)$. Now, let's focus on the part in the parentheses: $x^4 - 4x^2 - 5$. This looks a bit like a quadratic equation, right? Remember quadratics, those are usually in the form $ax^2 + bx + c$. Our expression has $x^4$ and $x^2$. This is what we call a quadratic form. We can treat it like a quadratic by making a substitution. Let $u = x^2$. If $u = x^2$, then $u^2 = (x2)2 = x^4$. So, we can rewrite our expression $x^4 - 4x^2 - 5$ as $u^2 - 4u - 5$.

Now, this looks like a standard quadratic that we can factor! We're looking for two numbers that multiply to -5 and add up to -4. Let's think about the factors of -5. They could be (1, -5) or (-1, 5).

  • If we try (1, -5): $1 \times -5 = -5$ and $1 + (-5) = -4$. Bingo! These are our numbers.

So, we can factor $u^2 - 4u - 5$ as $(u + 1)(u - 5)$.

But wait, we're not done yet! Remember, we made a substitution: $u = x^2$. We need to substitute back to get our answer in terms of 'x'. So, we replace 'u' with $x^2$ in our factored form:

(u+1)(u−5)ightarrow(x2+1)(x2−5) (u + 1)(u - 5) ightarrow (x^2 + 1)(x^2 - 5)

This is a crucial step, guys. Don't forget to substitute back! It's like solving a puzzle, and each piece needs to fit perfectly. So far, our original expression is factored as $y(x^2 + 1)(x^2 - 5)$. We're getting closer to the final answer, but we need to make sure it's completely factored. Let's move on to the final check.

Step 3: Checking for Further Factoring

We've successfully factored $x^4 y - 4x^2 y - 5y$ into $y(x^2 + 1)(x^2 - 5)$. Now, the final boss battle: is this expression completely factored? This means we need to check if any of the factors we have can be factored further. Let's examine each factor individually.

First, we have the factor 'y'. This is just a single variable, so it can't be factored any more. It's as simple as it gets.

Next, we have the factor $(x^2 + 1)$. Can we factor this any further using real numbers? This is a sum of squares. Generally, a sum of squares like $a^2 + b^2$ cannot be factored into simpler binomials with real coefficients. If we were working with complex numbers, we could factor it as $(x + i)(x - i)$, but in standard algebra problems, we usually stick to real factors. So, $(x^2 + 1)$ is considered prime (or irreducible) over the real numbers.

Finally, we have the factor $(x^2 - 5)$. Can this be factored further? This looks like a difference of squares, but 5 is not a perfect square. However, we can think of 5 as $\sqrt{5}^2$. So, this is in the form $a^2 - b^2$, where $a = x$ and $b = \sqrt{5}$. The difference of squares pattern tells us that $a^2 - b^2 = (a + b)(a - b)$. Applying this here, we get:

x2−5=(x+5)(x−5) x^2 - 5 = (x + \sqrt{5})(x - \sqrt{5})

So, it seems like we can factor $(x^2 - 5)$ further! This means our expression $y(x^2 + 1)(x^2 - 5)$ is not yet the completely factored form. The completely factored form would be $y(x^2 + 1)(x + \sqrt{5})(x - \sqrt{5})$.

However, there's a common convention in algebra problems, especially those presented in multiple-choice formats like the one you provided, where