Unlocking Rectangle Width: A Math Problem Explained

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Hey math enthusiasts! Ever found yourself scratching your head over a geometry problem? Today, we're diving into a classic: finding the width of a rectangle when we know its area and length. Sounds tricky? Don't worry, we'll break it down step by step, making it super easy to understand. We'll be using the area of a rectangle is (x4+4x3+3x2βˆ’4xβˆ’4)\left(x^4+4 x^3+3 x^2-4 x-4\right), and the length of the rectangle is (x3+5x2+8x+4)\left(x^3+5 x^2+8 x+4\right). If area = length Γ— width, what is the width of the rectangle?

Understanding the Basics: Area, Length, and Width

Alright, let's start with the basics. Remember those shapes we learned about in school? Well, a rectangle is a four-sided shape where opposite sides are equal, and all angles are right angles (90 degrees). Now, the area of a rectangle is the space it covers – imagine it as the amount of carpet needed to cover the floor. We calculate the area by multiplying the length (the longer side) by the width (the shorter side). So, the golden rule here is: Area = Length Γ— Width. Got it? Great!

Now, let's talk about the problem we're facing. We're given the area, which is (x4+4x3+3x2βˆ’4xβˆ’4)\left(x^4+4 x^3+3 x^2-4 x-4\right), and the length, which is (x3+5x2+8x+4)\left(x^3+5 x^2+8 x+4\right). Our mission? To find the width. This kind of problem often pops up in algebra, and it's all about using what we know to find what we don't. Think of it like a puzzle – we have some pieces (the area and length) and need to find the missing one (the width). Don't worry, we will find this piece, It's like a treasure hunt, but with math!

This isn't just about formulas; it's about seeing how math applies to real-world problems. Whether you're a student, a math lover, or someone who just likes to keep their mind sharp, this is for you. We'll break down the concepts to ensure that you'll have a clear grasp of them. Understanding area and how it relates to length and width is a foundational skill in geometry and algebra. We'll show you how to manipulate equations and solve for an unknown variable (the width), which is an important skill in problem-solving. So, let's roll up our sleeves and get into it. Prepare for an exciting journey through the world of rectangles and algebraic expressions.

Setting Up the Equation: The Key to Solving

Okay, guys, let's get down to business. We know that Area = Length Γ— Width. We can rearrange this to solve for the width, like so: Width = Area / Length. This is our main equation. Think of it as the roadmap to our answer. Now we have all we need to get started. Let's start with writing down what we know:

  • Area = (x4+4x3+3x2βˆ’4xβˆ’4)\left(x^4+4 x^3+3 x^2-4 x-4\right)
  • Length = (x3+5x2+8x+4)\left(x^3+5 x^2+8 x+4\right)

Our job is to divide the area by the length. When we divide a polynomial by another polynomial, we use a method called polynomial long division. It's similar to the long division you learned in elementary school, but with variables and exponents. Sounds daunting? It's not as bad as it seems, let's walk through it together.

The essential skill here is the ability to break down complex expressions into simpler forms. We are going to use the division formula, and this means we'll need to divide the area polynomial by the length polynomial. The key is to pay close attention to the exponents and coefficients. The reason why this is important, is because it helps us to manage complex calculations more easily. It's a structured approach that ensures we don't miss any steps, making it much easier to solve the problem and reduces the chance of making a mistake. It's a fundamental part of the problem-solving strategy and sets the stage for success.

Ready to get started? We will use polynomial long division. Let's make it easier. We will start with the first term of the dividend (the area) and divide it by the first term of the divisor (the length). This will give us the first term of our quotient (the width). Then, we will take the answer and multiply it by the length, and we will subtract the result from the area. This gives us the new polynomial. We will repeat these steps until there is nothing left. Are you ready?

Polynomial Long Division: The Heart of the Solution

Alright, let's dive into the core of the problem: polynomial long division. This is where we take the area ((x4+4x3+3x2βˆ’4xβˆ’4)\left(x^4+4 x^3+3 x^2-4 x-4\right)) and divide it by the length ((x3+5x2+8x+4)\left(x^3+5 x^2+8 x+4\right)). It looks a little intimidating at first, but trust me, it's manageable. We're going to break it down into simple steps.

  1. Set up the division: Write down the area inside the division symbol and the length outside.

        ___________
    

x3+5x2+8x+4 | x4+4x3+3x^2-4x-4 ```

  1. Divide the first terms: Divide the first term of the area (x4x^4) by the first term of the length (x3x^3). x4/x3=xx^4 / x^3 = x. So, we write 'x' at the top of the division.

            x
    

x3+5x2+8x+4 | x4+4x3+3x^2-4x-4 ```

  1. Multiply: Multiply 'x' (the result) by the entire length (x3+5x2+8x+4x^3 + 5x^2 + 8x + 4). This gives us x4+5x3+8x2+4xx^4 + 5x^3 + 8x^2 + 4x.

            x
    

x3+5x2+8x+4 | x4+4x3+3x^2-4x-4 x4+5x3+8x^2+4x ```

  1. Subtract: Subtract the result from the area.

            x
    

x3+5x2+8x+4 | x4+4x3+3x^2-4x-4 x4+5x3+8x^2+4x --------- -x3-5x2-8x-4 ```

  1. Bring down: Bring down the next term from the area.

            x
    

x3+5x2+8x+4 | x4+4x3+3x^2-4x-4 x4+5x3+8x^2+4x --------- -x3-5x2-8x-4 ```

  1. Repeat: Now, we repeat the process. Divide the first term of the new polynomial (βˆ’x3-x^3) by the first term of the length (x3x^3). βˆ’x3/x3=βˆ’1-x^3 / x^3 = -1. Write '-1' at the top.

            x-1
    

x3+5x2+8x+4 | x4+4x3+3x^2-4x-4 x4+5x3+8x^2+4x --------- -x3-2x2-8x-4 -x3-5x2-8x-4 --------- 0 ```

So, after all that hard work, the width is xβˆ’1x - 1. This means the correct answer is option D.

Checking the Answer: Always a Good Idea

  • Verify the Solution: To make sure we've got it right, we can multiply the length by the width and see if it gives us the area. That is, check that (x3+5x2+8x+4)(xβˆ’1)=x4+4x3+3x2βˆ’4xβˆ’4(x^3 + 5x^2 + 8x + 4)(x - 1) = x^4 + 4x^3 + 3x^2 - 4x - 4. If it does, we know we're golden.

    Let's expand the expression: (x3+5x2+8x+4)(xβˆ’1)=x4+5x3+8x2+4xβˆ’x3βˆ’5x2βˆ’8xβˆ’4(x^3 + 5x^2 + 8x + 4)(x - 1) = x^4 + 5x^3 + 8x^2 + 4x - x^3 - 5x^2 - 8x - 4.

    Combining like terms: x4+(5x3βˆ’x3)+(8x2βˆ’5x2)+(4xβˆ’8x)βˆ’4=x4+4x3+3x2βˆ’4xβˆ’4x^4 + (5x^3 - x^3) + (8x^2 - 5x^2) + (4x - 8x) - 4 = x^4 + 4x^3 + 3x^2 - 4x - 4

    Since this matches our given area, our answer is correct!

Conclusion: Width Found!

Alright, folks, we did it! We successfully found the width of the rectangle. By understanding the relationship between area, length, and width, and using polynomial long division, we cracked the code. Always remember, in math, it's not just about getting the answer; it's about the journey and the skills you learn along the way. Keep practicing, keep exploring, and keep the math adventures going! Thanks for joining me on this math journey. Until next time, keep calculating!