Simplify Polynomial Fractions: F(x)/g(x)

When you're diving into the world of algebra, you'll often encounter situations where you need to simplify expressions involving functions. One common task is dividing one polynomial by another. Today, we're going to break down how to find $\frac{f(x)}{g(x)}$ given $f(x)=16 x^5-48 x^4-8 x^3$ and $g(x)=8 x^2$. This process is fundamental to understanding more complex algebraic manipulations and is a stepping stone to concepts like rational functions. Don't worry if it seems a bit daunting at first; we'll go through it step-by-step, making sure you grasp each part of the process. The key here is to remember the rules of exponents and how to factor out common terms. It's like solving a puzzle, where each piece (each term in the polynomial) needs to fit correctly to get the final, simplified answer. We'll be using basic division principles that you likely learned in earlier math courses, applied specifically to algebraic expressions. So, get ready to flex those math muscles and simplify this expression with us!

Understanding the Functions

Before we start dividing, let's take a moment to understand the two functions we're working with: $f(x)=16 x^5-48 x^4-8 x^3$ and $g(x)=8 x^2$. The function $f(x)$ is a polynomial of degree 5, meaning the highest power of $x$ in the expression is 5. It has three terms: $16x^5$, $-48x^4$, and $-8x^3$. The function $g(x)$ is a simpler polynomial, specifically a monomial (a polynomial with only one term), of degree 2. Our goal is to compute the division $\frac{f(x)}{g(x)}$, which means we want to divide the entire expression of $f(x)$ by the entire expression of $g(x)$. This operation essentially asks us to simplify the ratio of these two polynomials. It’s important to recognize that we can only perform this division if $g(x)$ is not equal to zero. In this case, $g(x) = 8x^2$, which is zero only when $x=0$. Therefore, our simplified expression will be valid for all $x \neq 0$. When we're simplifying a fraction, whether it's with numbers or polynomials, we're looking for common factors that can be canceled out. Think of it like reducing a fraction, say $\frac{10}{15}$. You know that both 10 and 15 are divisible by 5, so you divide both by 5 to get $\frac{2}{3}$. The same principle applies here, but instead of just numbers, we'll be dealing with coefficients (the numbers in front of the variables) and the variables themselves raised to different powers. We'll need to apply the rules of exponents, specifically when dividing powers of the same base, which states that $x^a / x^b = x^{a-b}$. This rule will be crucial in simplifying each term of the polynomial division. Let's prepare to apply these rules to our specific functions.

Step-by-Step Division Process

Now, let's get down to the actual process of finding $\frac{f(x)}{g(x)}$. We have $f(x) = 16x^5 - 48x^4 - 8x^3$ and $g(x) = 8x^2$. To find $\frac{f(x)}{g(x)}$, we write it out as:

$$\frac{16x^5 - 48x^4 - 8x^3}{8x^2}$$

Since $g(x)$ is a monomial, we can divide each term of $f(x)$ by $g(x)$ separately. This is a key simplification strategy when the denominator is a single term. So, we can rewrite the expression as:

$$\frac{16x^5}{8x^2} - \frac{48x^4}{8x^2} - \frac{8x^3}{8x^2}$$

Now, we tackle each fraction individually. Let's start with the first term: $\frac{16x^5}{8x^2}$. To simplify this, we divide the coefficients (16 by 8) and then divide the variables using the rule of exponents ($x^a / x^b = x^{a-b}$). So, $\frac{16}{8} = 2$, and $\frac{x^5}{x^2} = x^{5-2} = x^3$. Combining these, the first term simplifies to $2x^3$.

Next, let's look at the second term: $\frac{-48x^4}{8x^2}$. Again, we divide the coefficients: $\frac{-48}{8} = -6$. For the variables, we have $\frac{x^4}{x^2} = x^{4-2} = x^2$. So, the second term simplifies to $-6x^2$.

Finally, we have the third term: $\frac{-8x^3}{8x^2}$. Dividing the coefficients gives us $\frac{-8}{8} = -1$. For the variables, we have $\frac{x^3}{x^2} = x^{3-2} = x^1$, which is simply $x$. So, the third term simplifies to $-1x$, or just $-x$.

Putting all the simplified terms back together, we get our final answer for $\frac{f(x)}{g(x)}$:

$$2x^3 - 6x^2 - x$$

This is the simplified form of the original fraction. It's important to remember that this result is valid for all values of $x$ except $x=0$, because division by zero is undefined. This step-by-step approach, breaking down the complex expression into simpler parts, is a powerful technique in algebra. It allows us to manage complexity by applying basic arithmetic and algebraic rules to each component.

Factoring for Simplification (Alternative Approach)

Another way to approach the simplification of $\frac{f(x)}{g(x)}$ is by factoring. While the direct division method is straightforward when the denominator is a monomial, factoring can be useful for understanding the underlying structure of the expression and is essential for more complex rational functions. Let's revisit our functions: $f(x)=16 x^5-48 x^4-8 x^3$ and $g(x)=8 x^2$. Our goal is to find $\frac{f(x)}{g(x)}$.

First, let's look at the numerator, $f(x) = 16x^5 - 48x^4 - 8x^3$. We need to find the greatest common factor (GCF) of all the terms. Let's consider the coefficients first: 16, -48, and -8. The greatest common divisor of these numbers is 8. Now, let's look at the variables: $x^5$, $x^4$, and $x^3$. The lowest power of $x$ present in all terms is $x^3$. Therefore, the GCF of $f(x)$ is $8x^3$.

We can factor out $8x^3$ from each term in $f(x)$:

• $16x^5 = (8x^3) \cdot (2x^2)$
• $-48x^4 = (8x^3) \cdot (-6x)$
• $-8x^3 = (8x^3) \cdot (-1)$

So, we can rewrite $f(x)$ as:

$$f(x) = 8x^3(2x^2 - 6x - 1)$$

Now, let's look at the denominator, $g(x) = 8x^2$. We can also think of this as a factorized form, as it's already a single term.

Now, we can substitute the factored form of $f(x)$ back into our fraction:

$$\frac{f(x)}{g(x)} = \frac{8x^3(2x^2 - 6x - 1)}{8x^2}$$

We can see that both the numerator and the denominator have common factors. The number 8 is a common factor. The variable part has $x^3$ in the numerator and $x^2$ in the denominator. We can cancel out $8$ from both. For the variables, we can cancel $x^2$ from both the numerator and the denominator. Remember that $\frac{x^3}{x^2} = x^{3-2} = x^1 = x$.

So, after canceling out the common factors, we are left with:

$$\frac{\cancel{8}x^{\cancel{3}} (2x^2 - 6x - 1)}{\cancel{8}x^{\cancel{2}}} = x(2x^2 - 6x - 1)$$

Now, we can distribute the remaining $x$ back into the parenthesis to get the final simplified expression:

$$x \cdot 2x^2 - x \cdot 6x - x \cdot 1$$

This results in:

$$2x^3 - 6x^2 - x$$

This matches the result we obtained using the direct division method. Factoring often provides a clearer view of what can be canceled and why. It reinforces the understanding of algebraic structures and is a vital skill for tackling more advanced topics in mathematics, such as partial fraction decomposition and the analysis of rational functions. Both methods, direct division and factoring, lead to the same correct answer, confirming our understanding and ability to manipulate these algebraic expressions effectively. This dual approach also highlights how different algebraic techniques can be used to solve the same problem, offering flexibility and deeper insight.

Conclusion and Further Exploration

We have successfully found $\frac{f(x)}{g(x)}$ for $f(x)=16 x^5-48 x^4-8 x^3$ and $g(x)=8 x^2$. Using two different methods – direct division of each term and factoring out the greatest common factor – we arrived at the same simplified expression: $2x^3 - 6x^2 - x$. This demonstrates the robustness of algebraic principles and provides confidence in our ability to simplify polynomial fractions. Remember, the key steps involve dividing coefficients and subtracting exponents when dividing powers of the same base, or finding and canceling common factors between the numerator and denominator. It's crucial to note that this simplification is valid for all $x \neq 0$, as division by zero is undefined. Understanding this process is foundational for many areas of mathematics. For instance, when you move on to study rational functions, which are ratios of two polynomials, these simplification techniques become even more critical. They help in graphing these functions, finding their asymptotes, and analyzing their behavior.

If you're interested in exploring more about polynomial division, rational functions, and their applications, I highly recommend checking out resources that delve deeper into these topics. For a comprehensive understanding of algebraic concepts and further practice, the Khan Academy website offers excellent tutorials and exercises that cover everything from basic algebra to advanced calculus. Their explanations are clear, and their practice problems are invaluable for solidifying your knowledge. Additionally, exploring Paul's Online Math Notes can provide detailed explanations and worked examples on various calculus and algebra topics, serving as a fantastic reference for students at all levels.