Factoring A Cubic Polynomial: A Step-by-Step Guide

by Andrew McMorgan 51 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of algebra and tackling a question that might seem a little tricky at first glance: What is the factored form of 2x3+4x2βˆ’x2x^3 + 4x^2 - x? Don't sweat it if you're not a math wizard yet; we're going to break this down piece by piece, making sure everyone can follow along. Finding the factored form of a polynomial is like unlocking a secret code, revealing the building blocks of that expression. It's a super useful skill in mathematics, especially when you're trying to solve equations, graph functions, or simplify complex expressions. So, grab your favorite beverage, get comfy, and let's get this algebra party started!

Understanding Polynomials and Factoring

Before we jump into solving our specific problem, let's quickly chat about what polynomials and factoring actually are. A polynomial is basically a mathematical expression made up of variables (like 'x' in our case) and coefficients, combined using addition, subtraction, multiplication, and non-negative integer exponents. Think of it as a sum of terms, where each term is a number multiplied by a variable raised to a power. Our expression, 2x3+4x2βˆ’x2x^3 + 4x^2 - x, is a cubic polynomial because the highest power of 'x' is 3.

Now, factoring is the reverse of expanding. When you expand an expression, you multiply factors together to get a larger expression. When you factor, you start with the larger expression and break it down into its original multiplicative components, its factors. For instance, if you have the number 12, its factors are 1, 2, 3, 4, 6, and 12. Factoring a polynomial means finding simpler polynomials that, when multiplied together, give you the original polynomial. This is super important for solving polynomial equations because if a product of factors equals zero, then at least one of the factors must be zero. This is the Zero Product Property, and it's a game-changer!

So, when we're asked to find the factored form of 2x3+4x2βˆ’x2x^3 + 4x^2 - x, we're looking for a way to rewrite this expression as a product of simpler expressions. The options provided – A, B, C, and D – all represent potential factored forms. Our job is to figure out which one is the correct one by either factoring the original polynomial ourselves or by expanding each of the options to see which one matches the original expression.

Step-by-Step Factoring: Finding the Greatest Common Factor (GCF)

Alright, let's get down to business with our expression: 2x3+4x2βˆ’x2x^3 + 4x^2 - x. The first and most crucial step in factoring any polynomial is to look for a Greatest Common Factor (GCF). The GCF is the largest factor that all the terms in the polynomial share. To find it, we examine the coefficients (the numbers) and the variables in each term.

Our terms are 2x32x^3, 4x24x^2, and βˆ’x-x. Let's look at the coefficients first: 2, 4, and -1. The greatest common factor of 2, 4, and 1 is simply 1. So, the numerical part of our GCF is 1.

Now, let's look at the variables: x3x^3, x2x^2, and xx. All terms have 'x' in them. To find the GCF of the variables, we take the variable with the lowest exponent. In this case, the lowest exponent of 'x' is 1 (from the βˆ’x-x term, which can be thought of as βˆ’1x1-1x^1). So, the variable part of our GCF is xx.

Combining the numerical and variable parts, the GCF of 2x3+4x2βˆ’x2x^3 + 4x^2 - x is xx. This is awesome because it means we can factor out an 'x' from every term.

When we factor out 'x', we're essentially dividing each term by 'x' and placing 'x' outside parentheses. Let's see what happens:

  • For the first term, 2x32x^3: rac{2x^3}{x} = 2x^{3-1} = 2x^2
  • For the second term, 4x24x^2: rac{4x^2}{x} = 4x^{2-1} = 4x^1 = 4x
  • For the third term, βˆ’x-x: rac{-x}{x} = -1

So, when we factor out xx from 2x3+4x2βˆ’x2x^3 + 4x^2 - x, we get: x(2x2+4xβˆ’1)x(2x^2 + 4x - 1).

This looks promising, right? This expression, x(2x2+4xβˆ’1)x(2x^2 + 4x - 1), is now in a factored form. The part inside the parentheses, 2x2+4xβˆ’12x^2 + 4x - 1, is a quadratic expression. We could try to factor this quadratic further, but for now, let's just focus on whether our factored form matches any of the given options.

Looking at the options:

  • A. 2x(x2+2x+1)2x(x^2+2x+1)
  • B. x(2x2+4x+1)x(2x^2+4x+1)
  • C. 2x(x2+2xβˆ’1)2x(x^2+2x-1)
  • D. x(2x2+4xβˆ’1)x(2x^2+4x-1)

Holy moly, option D is exactly what we got! x(2x2+4xβˆ’1)x(2x^2 + 4x - 1). This means we've likely found our answer. But, to be absolutely sure, especially in a test scenario, it's always a good idea to double-check our work. We can do this by expanding option D to see if we get back our original polynomial.

Verifying the Answer: Expanding the Factored Form

Our potential answer is x(2x2+4xβˆ’1)x(2x^2 + 4x - 1). To verify if this is indeed the correct factored form of 2x3+4x2βˆ’x2x^3 + 4x^2 - x, we'll use the distributive property. This means we multiply the 'x' outside the parentheses by each term inside the parentheses.

Let's do it:

  • Multiply xx by 2x22x^2: ximes2x2=2x1+2=2x3x imes 2x^2 = 2x^{1+2} = 2x^3
  • Multiply xx by 4x4x: ximes4x=4x1+1=4x2x imes 4x = 4x^{1+1} = 4x^2
  • Multiply xx by βˆ’1-1: ximes(βˆ’1)=βˆ’xx imes (-1) = -x

Now, we combine these results:

x(2x2+4xβˆ’1)=2x3+4x2βˆ’xx(2x^2 + 4x - 1) = 2x^3 + 4x^2 - x

Boom! We got our original polynomial back. This confirms that x(2x2+4xβˆ’1)x(2x^2 + 4x - 1) is indeed the correct factored form. So, option D is our winner, guys!

Why Other Options Are Incorrect

It's also super helpful to understand why the other options don't work. This reinforces our understanding and helps us avoid common mistakes. Let's quickly expand options A, B, and C to see why they don't match 2x3+4x2βˆ’x2x^3 + 4x^2 - x.

Option A: 2x(x2+2x+1)2x(x^2+2x+1)

Let's distribute the 2x2x:

  • 2ximesx2=2x1+2=2x32x imes x^2 = 2x^{1+2} = 2x^3
  • 2ximes2x=4x1+1=4x22x imes 2x = 4x^{1+1} = 4x^2
  • 2ximes1=2x2x imes 1 = 2x

Combining these, we get: 2x3+4x2+2x2x^3 + 4x^2 + 2x. This is not 2x3+4x2βˆ’x2x^3 + 4x^2 - x because of the last term (+2x+2x instead of βˆ’x-x). The mistake here is likely factoring out 2x2x instead of xx, and also the sign in the original expression was ignored.

Option B: x(2x2+4x+1)x(2x^2+4x+1)

Let's distribute the xx:

  • ximes2x2=2x1+2=2x3x imes 2x^2 = 2x^{1+2} = 2x^3
  • ximes4x=4x1+1=4x2x imes 4x = 4x^{1+1} = 4x^2
  • ximes1=xx imes 1 = x

Combining these, we get: 2x3+4x2+x2x^3 + 4x^2 + x. This is also not 2x3+4x2βˆ’x2x^3 + 4x^2 - x. The error here is the sign of the constant term. The original polynomial has a βˆ’x-x term, but this option ends up with a +x+x term. This highlights the importance of carefully handling signs during factoring.

Option C: 2x(x2+2xβˆ’1)2x(x^2+2x-1)

Let's distribute the 2x2x:

  • 2ximesx2=2x1+2=2x32x imes x^2 = 2x^{1+2} = 2x^3
  • 2ximes2x=4x1+1=4x22x imes 2x = 4x^{1+1} = 4x^2
  • 2ximes(βˆ’1)=βˆ’2x2x imes (-1) = -2x

Combining these, we get: 2x3+4x2βˆ’2x2x^3 + 4x^2 - 2x. Again, this is not 2x3+4x2βˆ’x2x^3 + 4x^2 - x. The issue here is twofold: first, factoring out 2x2x when the GCF is just xx, and second, the final term is βˆ’2x-2x instead of βˆ’x-x. This often happens if you miscalculate the GCF or misapply the division.

The Power of Factoring: Beyond This Problem

So there you have it, guys! The factored form of 2x3+4x2βˆ’x2x^3 + 4x^2 - x is x(2x2+4xβˆ’1)x(2x^2 + 4x - 1). We found this by first identifying the Greatest Common Factor (GCF), which in this case was xx, and then dividing each term of the original polynomial by the GCF. We then confirmed our answer by expanding the factored form, ensuring it matched the original expression. Understanding how to factor polynomials is a fundamental skill in algebra that opens doors to solving more complex problems.

Think about it: if we wanted to solve the equation 2x3+4x2βˆ’x=02x^3 + 4x^2 - x = 0, we could rewrite it as x(2x2+4xβˆ’1)=0x(2x^2 + 4x - 1) = 0. Using the Zero Product Property, we know that either x=0x = 0 or 2x2+4xβˆ’1=02x^2 + 4x - 1 = 0. The first solution is super easy! For the quadratic part, we might need other techniques like the quadratic formula if it doesn't factor easily, but we've already simplified the problem significantly by factoring out the common term.

Factoring also helps in simplifying rational expressions (fractions with polynomials), graphing polynomials by revealing their roots (where the graph crosses the x-axis), and understanding the behavior of functions. It's a cornerstone concept that you'll build upon as you continue your math journey. Don't be discouraged if it takes practice; every mathematician started by learning these basic steps. Keep practicing, keep asking questions, and remember that breaking down complex problems into smaller, manageable steps is the key to success. Until next time, keep those minds sharp!