Polynomial Division (2x^2-15x+18) Divided By (2x+1)

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Hey guys! Today, we're diving into the world of polynomial division, and we're going to tackle a specific problem: dividing 2x2βˆ’15x+18{ 2x^2 - 15x + 18 } by 2x+1{ 2x + 1 }. Polynomial division might seem intimidating at first, but trust me, once you get the hang of it, it's not that bad. We’ll break it down step by step, so you can confidently solve similar problems. Let's get started!

Understanding Polynomial Division

Before we jump into the specific problem, let’s take a moment to understand what polynomial division actually means. Think of it like regular long division, but instead of numbers, we're working with expressions that include variables and exponents. The goal is the same: to find out how many times one polynomial fits into another and what's left over (the remainder). Polynomial division is a fundamental operation in algebra, and mastering it opens doors to more advanced topics like factoring, solving equations, and graphing functions. It's super useful for simplifying complex expressions and understanding the behavior of polynomial functions. This concept is especially crucial when dealing with rational functions, where you often need to divide polynomials to simplify or analyze the function. So, understanding the underlying principles will help you a lot in the long run. For instance, if you're working with a rational expression and need to find its asymptotes, polynomial division might be a necessary step. Similarly, when you're trying to solve a polynomial equation, dividing by a known factor can help you reduce the degree of the polynomial and make it easier to find the roots. So, whether you're into calculus, pre-calculus, or just algebra, polynomial division is a skill you'll use frequently.

The Long Division Method

The most common method for polynomial division is long division, which is very similar to the long division you learned in elementary school. Let's quickly recap the steps involved:

  1. Set up the division: Write the dividend (the polynomial being divided) inside the division symbol and the divisor (the polynomial you're dividing by) outside.
  2. Divide the leading terms: Divide the leading term of the dividend by the leading term of the divisor. This gives you the first term of the quotient.
  3. Multiply: Multiply the entire divisor by the first term of the quotient.
  4. Subtract: Subtract the result from the dividend.
  5. Bring down: Bring down the next term of the dividend.
  6. Repeat: Repeat steps 2-5 until there are no more terms to bring down.
  7. Remainder: The polynomial left over at the end is the remainder.

Why is long division so important? Well, it provides a systematic way to break down a complex division problem into smaller, manageable steps. It ensures that you account for each term and exponent correctly, minimizing the chances of making mistakes. Plus, it gives you a clear picture of the quotient and the remainder, which are both crucial pieces of information. Another reason why long division is so valuable is its versatility. It works for any polynomial division problem, regardless of the degree or complexity of the polynomials involved. Whether you're dividing a quadratic by a linear term or a higher-degree polynomial by a quadratic, the long division method remains consistent and reliable.

Step-by-Step Solution for (2x2βˆ’15x+18)Γ·(2x+1){ (2x^2 - 15x + 18) \div (2x + 1) }

Now that we've got the basics covered, let's tackle our specific problem: (2x2βˆ’15x+18)Γ·(2x+1){ (2x^2 - 15x + 18) \div (2x + 1) }. We'll walk through each step of the long division process to find the quotient and remainder.

1. Set Up the Division

First, write the dividend (2x2βˆ’15x+18){ (2x^2 - 15x + 18) } inside the division symbol and the divisor (2x+1){ (2x + 1) } outside. This sets up the framework for our long division process. Ensuring you align the terms correctly by their degrees (powers of x) is crucial for avoiding errors later on. Think of it like setting the stage for a play; a well-organized setup makes the performance (the division process) run smoothly.

2. Divide the Leading Terms

Next, we're going to focus on the leading terms: 2x2{ 2x^2 } from the dividend and 2x{ 2x } from the divisor. Divide 2x2{ 2x^2 } by 2x{ 2x }, which gives us x{ x }. This x{ x } is the first term of our quotient. This step is the heart of polynomial division – figuring out what term to multiply the divisor by to match the highest-degree term of the dividend. It's like finding the perfect piece to fit into a puzzle. Make sure you're dividing the coefficients and subtracting the exponents correctly (remember, x2Γ·x=x2βˆ’1=x{ x^2 \div x = x^{2-1} = x }). This first term sets the stage for the rest of the division, so accuracy here is key.

3. Multiply

Now, multiply the entire divisor (2x+1){ (2x + 1) } by the first term of the quotient, which is x{ x }. So, x(2x+1)=2x2+x{ x(2x + 1) = 2x^2 + x }. This step is where we distribute the first term of the quotient across the entire divisor. It's crucial to multiply x{ x } by both 2x{ 2x } and 1{ 1 }, ensuring you get both terms of the product. Think of it like expanding a bracket in algebra. A common mistake here is forgetting to multiply x{ x } by both terms, so double-check your work.

4. Subtract

Subtract the result (2x2+x){ (2x^2 + x) } from the corresponding terms in the dividend (2x2βˆ’15x+18){ (2x^2 - 15x + 18) }. This means (2x2βˆ’15x)βˆ’(2x2+x){ (2x^2 - 15x) - (2x^2 + x) }. Remember to distribute the negative sign correctly: 2x2βˆ’15xβˆ’2x2βˆ’x=βˆ’16x{ 2x^2 - 15x - 2x^2 - x = -16x }. The 2x2{ 2x^2 } terms cancel out, which is exactly what we want. This subtraction step is where we eliminate the leading term of the dividend, making the polynomial smaller and easier to work with. Accuracy in this step is crucial because any mistake here will propagate through the rest of the division. Be especially careful with the signs – subtracting a positive term becomes negative, and subtracting a negative term becomes positive.

5. Bring Down

Bring down the next term from the dividend, which is +18{ +18 }. This gives us the new expression βˆ’16x+18{ -16x + 18 }. This step is similar to bringing down the next digit in regular long division. It essentially adds the next piece of the puzzle to our working polynomial. By bringing down the +18{ +18 }, we're setting up the next iteration of the division process. Make sure you bring down the term with the correct sign.

6. Repeat

Repeat the process: divide the leading term of the new expression βˆ’16x{ -16x } by the leading term of the divisor 2x{ 2x }. This gives us βˆ’8{ -8 }, which is the next term of our quotient. Just like before, we're focusing on the leading terms to figure out what to multiply the divisor by. Make sure you include the sign in your division – in this case, we're dividing βˆ’16x{ -16x } by 2x{ 2x }, which results in a negative quotient term, βˆ’8{ -8 }. This sign is crucial for the subsequent steps.

7. Multiply Again

Multiply the entire divisor (2x+1){ (2x + 1) } by βˆ’8{ -8 }: βˆ’8(2x+1)=βˆ’16xβˆ’8{ -8(2x + 1) = -16x - 8 }. Again, we're distributing the new quotient term across the entire divisor. Make sure you multiply βˆ’8{ -8 } by both 2x{ 2x } and 1{ 1 }, paying close attention to the signs. A common mistake is forgetting to multiply βˆ’8{ -8 } by both terms, so double-check your work to avoid errors.

8. Subtract Again

Subtract the result (βˆ’16xβˆ’8){ (-16x - 8) } from the current expression (βˆ’16x+18){ (-16x + 18) }: (βˆ’16x+18)βˆ’(βˆ’16xβˆ’8){ (-16x + 18) - (-16x - 8) }. Distribute the negative sign carefully: βˆ’16x+18+16x+8=26{ -16x + 18 + 16x + 8 = 26 }. Notice that the βˆ’16x{ -16x } terms cancel out, leaving us with a constant term. This step is crucial for simplifying the expression and revealing the remainder. As before, be extra cautious with the signs – subtracting a negative term becomes positive.

9. Remainder

Since there are no more terms to bring down and the degree of 26{ 26 } (which is 0) is less than the degree of 2x+1{ 2x + 1 } (which is 1), we've reached the end of the division. The remainder is 26{ 26 }. This remainder is what's