Polynomial Division: Find The Remainder R(x)

Aug 4, 2025 by ADMIN 45 views

Polynomial Division: Finding the Remainder When Dividing $f(x)$ by $d(x)$

Hey guys! Ever stumbled upon a polynomial division problem that looks like a beast? Fear not! We're going to break down how to tackle these, especially when the focus is on finding that sneaky remainder, $R(x)$ . Let's dive into an example that will make polynomial division feel like a piece of cake. We'll focus on the specific case of dividing $f(x)$ by $d(x)$ and pinpointing that $R(x)$ term. This guide is designed to help you not just solve the problem but truly understand the mechanics behind polynomial division.

Understanding the Basics of Polynomial Division

Before we jump into the nitty-gritty, let's quickly recap what polynomial division is all about. Think of it like regular long division, but with polynomials! The goal is to divide a polynomial, $f(x)$ , by another polynomial, $d(x)$ , and express the result in a specific format:

$\frac{f(x)}{d(x)} = Q(x) + \frac{R(x)}{d(x)}$

Where:

$f(x)$ is the dividend (the polynomial being divided).
$d(x)$ is the divisor (the polynomial we're dividing by).
$Q(x)$ is the quotient (the result of the division).
$R(x)$ is the remainder (what's left over after the division).

Our main mission here is to find $R(x)$ , but understanding the whole process is key. So, we will cover all parts of the equation to make sure that you guys have a solid foundation on the topic. The beauty of this form is that it neatly separates the whole-polynomial part, $Q(x)$ , from the fractional part, which includes the remainder, $R(x)$ . This structure is crucial for various algebraic manipulations and is frequently used in calculus and other advanced mathematical contexts. When we talk about dividing polynomials, we're essentially asking: "How many times does $d(x)$ fit into $f(x)$ completely, and what's the leftover?" The quotient $Q(x)$ tells us how many times $d(x)$ fits into $f(x)$ , and the remainder $R(x)$ represents the part of $f(x)$ that $d(x)$ couldn't divide evenly.

The degree of the remainder $R(x)$ is always less than the degree of the divisor $d(x)$ . This is a fundamental property of polynomial division and a useful check to ensure your calculations are correct. If, for instance, you're dividing by a linear polynomial (degree 1), your remainder will be a constant (degree 0). If the degree of $R(x)$ were equal to or greater than the degree of $d(x)$ , it would imply that further division is possible. The process of polynomial division is not just a mechanical procedure; it's a way to decompose a complex polynomial expression into simpler parts. This decomposition can be incredibly useful for solving equations, simplifying expressions, and understanding the behavior of polynomial functions. The ability to divide polynomials efficiently and accurately is a foundational skill in algebra and is essential for success in higher-level mathematics courses.

Tackling the Specific Problem

Now, let's apply this to our specific problem:

$\frac{f(x)}{d(x)} = \frac{-2x^3 + 8x^2 + 12x - 7}{x - 5}$

Here, $f(x) = -2x^3 + 8x^2 + 12x - 7$ and $d(x) = x - 5$ . Our goal is to find $R(x)$ .

Step-by-Step Polynomial Long Division

We'll use polynomial long division, which is the most common method. It might seem a bit daunting at first, but trust me, it's a systematic process that gets easier with practice. This method mirrors the long division you learned with numbers, but now we're working with algebraic expressions.

Set up the division: Write the dividend ( $-2x^3 + 8x^2 + 12x - 7$ ) inside the division symbol and the divisor ( $x - 5$ ) outside.
```
            ________________________
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 ```

Divide the leading terms: Divide the leading term of the dividend ( $-2x^3$ ) by the leading term of the divisor ( $x$ ). This gives us $-2x^2$ . Write this above the division symbol, aligned with the $x^2$ term.
```
            -2x^2
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 ```

Multiply: Multiply the quotient term we just found ( $-2x^2$ ) by the entire divisor ( $x - 5$ ). This gives us $-2x^3 + 10x^2$ .
Subtract: Subtract the result from the corresponding terms in the dividend:
```
            -2x^2
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 -(-2x^3 + 10x^2) ____________________ -2x^2 ```

Bring down the next term: Bring down the next term from the dividend (+12x):
```
            -2x^2
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 -(-2x^3 + 10x^2) ____________________ -2x^2 + 12x ```

Repeat the process: Now, divide the leading term of the new expression ( $-2x^2$ ) by the leading term of the divisor ( $x$ ), which gives us $-2x$ . Write this in the quotient.
```
            -2x^2 - 2x
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 -(-2x^3 + 10x^2) ____________________ -2x^2 + 12x ```

Multiply: Multiply $-2x$ by $(x - 5)$ to get $-2x^2 + 10x$ .
Subtract: Subtract this from the current expression:
```
            -2x^2 - 2x
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 -(-2x^3 + 10x^2) ____________________ -2x^2 + 12x -(-2x^2 + 10x) ________________ 2x ```

Bring down the next term: Bring down the last term from the dividend (-7):
```
            -2x^2 - 2x
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 -(-2x^3 + 10x^2) ____________________ -2x^2 + 12x -(-2x^2 + 10x) ________________ 2x - 7 ```

Repeat one last time: Divide $2x$ by $x$ , which gives us $2$ . Write this in the quotient.
```
            -2x^2 - 2x + 2
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 -(-2x^3 + 10x^2) ____________________ -2x^2 + 12x -(-2x^2 + 10x) ________________ 2x - 7 ```

Multiply: Multiply $2$ by $(x - 5)$ to get $2x - 10$ .
Subtract: Subtract this from the current expression:
```
            -2x^2 - 2x + 2
```

x - 5 | -2x^3 + 8x^2 + 12x - 7 -(-2x^3 + 10x^2) ____________________ -2x^2 + 12x -(-2x^2 + 10x) ________________ 2x - 7 -(2x - 10) __________ 3 ```

Identify the remainder: The remainder is $3$ .

The Result

So, we've found that $R(x) = 3$ . This means that when we divide $-2x^3 + 8x^2 + 12x - 7$ by $x - 5$ , we get a quotient of $-2x^2 - 2x + 2$ and a remainder of $3$ .

We can express this as:

$\frac{-2x^3 + 8x^2 + 12x - 7}{x - 5} = -2x^2 - 2x + 2 + \frac{3}{x - 5}$

Focus on the Remainder: Why It Matters

Okay, so we found the remainder, but why is it so important? Well, the remainder theorem tells us that if we divide a polynomial $f(x)$ by $(x - a)$ , the remainder is equal to $f(a)$ . This is a super handy shortcut for evaluating polynomials at specific values. For instance, in our example, $d(x) = x - 5$ , so $a = 5$ . If we plug $5$ into $f(x)$ , we should get the remainder:

$f(5) = -2(5)^3 + 8(5)^2 + 12(5) - 7 = -250 + 200 + 60 - 7 = 3$

Yep, it matches our remainder! This is a great way to double-check your work and a powerful tool for polynomial analysis. Understanding the remainder also helps in factoring polynomials. If the remainder is zero, it means the divisor is a factor of the dividend. This is a cornerstone concept in algebra and is used extensively in solving polynomial equations.

The remainder also plays a critical role in calculus, particularly when dealing with rational functions (ratios of polynomials). Expressing a rational function in the form $Q(x) + \frac{R(x)}{d(x)}$ simplifies integration and other operations. The remainder theorem isn't just a theoretical concept; it has practical applications in various fields, including engineering, computer science, and economics, where polynomial models are used extensively.

Alternative Method: Synthetic Division

For those who love shortcuts, there's synthetic division! It's a faster way to divide polynomials, especially when the divisor is of the form $(x - a)$ . Let's see how it works with our problem. Synthetic division is particularly efficient because it focuses on the coefficients of the polynomials, streamlining the division process. It avoids the need to write out the variables, making it a more compact and less error-prone method once mastered.

Set up: Write down the coefficients of $f(x)$ : $-2$ , $8$ , $12$ , and $-7$ . Write the value of $a$ from the divisor $(x - 5)$ , which is $5$ , to the left.
```
5 | -2  8  12  -7
```
Bring down the first coefficient: Bring down the first coefficient ( $-2$ ) below the line.
```
5 | -2  8  12  -7
    -2
```
Multiply and add: Multiply the value we brought down ( $-2$ ) by $5$ , which gives $-10$ . Write this under the next coefficient ( $8$ ) and add them:
```
5 | -2  8  12  -7
       -10
    -2 -2
```
Repeat: Multiply $-2$ by $5$ to get $-10$ . Write this under $12$ and add them:
```
5 | -2  8  12  -7
       -10 -10
    -2 -2  2
```
Repeat again: Multiply $2$ by $5$ to get $10$ . Write this under $-7$ and add them:
```
5 | -2  8  12  -7
       -10 -10  10
    -2 -2  2   3
```
Interpret the result: The last number below the line ( $3$ ) is the remainder, $R(x)$ . The other numbers ( $-2$ , $-2$ , and $2$ ) are the coefficients of the quotient, $Q(x) = -2x^2 - 2x + 2$ .

See? We got the same remainder, $3$ , but with less writing! Synthetic division is particularly useful when dealing with higher-degree polynomials, as it can significantly reduce the complexity of the calculations. However, it's crucial to remember that synthetic division is only directly applicable when dividing by a linear divisor of the form $(x - a)$ . For divisors of higher degree, polynomial long division remains the go-to method.

Key Takeaways

Polynomial division helps us express $\frac{f(x)}{d(x)}$ as $Q(x) + \frac{R(x)}{d(x)}$ .
The remainder $R(x)$ is what's left over after division.
Polynomial long division is a reliable method for any divisor.
Synthetic division is a faster method for linear divisors $(x - a)$ .
The remainder theorem states that $R(x) = f(a)$ when dividing by $(x - a)$ .

Practice Makes Perfect

Polynomial division might seem tricky at first, but with practice, you'll become a pro! Try out different examples, and don't be afraid to make mistakes – that's how we learn! Keep practicing, and you'll find that polynomial division becomes second nature. Remember, the key is to understand the process, not just memorize the steps. By grasping the underlying concepts, you'll be able to tackle a wide range of polynomial problems with confidence.

So, to answer the original question:

$R(x) = 3$

And there you have it! You've successfully navigated polynomial division and found the remainder. Keep up the awesome work, guys! Remember, math is like a muscle; the more you exercise it, the stronger it gets. So, keep practicing, keep exploring, and most importantly, keep having fun with it! You've got this!