Factoring cubic polynomials can seem daunting, but with the right approach, it becomes manageable. This post explores proven methods to factorize the cubic equation x³ - 6x² + 11x - 6, providing a step-by-step guide you can apply to similar problems. We'll cover both the trial and error method and the rational root theorem, equipping you with versatile techniques for tackling cubic factorization.
Understanding the Problem: x³ - 6x² + 11x - 6
Before diving into the methods, let's understand what we're aiming for. Factoring this cubic polynomial means expressing it as a product of simpler expressions, ideally linear factors. Our goal is to find the values of 'x' that make the equation equal to zero, and those values will help us create the factors.
Method 1: Trial and Error (Integer Root Method)
This method relies on testing integer values of 'x' to see if they result in the polynomial equaling zero. If an integer value makes the equation zero (i.e., it's a root), then (x - that value) is a factor.
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Test Simple Integer Values: Let's start by trying simple integers like x = 1, x = 2, x = 3, etc.
- x = 1: 1³ - 6(1)² + 11(1) - 6 = 0. Success! This means (x - 1) is a factor.
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Polynomial Division: Now that we've found one factor, we use polynomial long division or synthetic division to divide the original cubic polynomial by (x - 1).
x² - 5x + 6 x - 1 | x³ - 6x² + 11x - 6Performing the division (you can use either long division or synthetic division), you will obtain the quotient x² - 5x + 6.
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Factor the Quadratic: The quotient x² - 5x + 6 is a quadratic that can be factored easily. We look for two numbers that add up to -5 and multiply to 6. These numbers are -2 and -3. Therefore:
x² - 5x + 6 = (x - 2)(x - 3)
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Final Factored Form: Combining our results, the completely factored form of x³ - 6x² + 11x - 6 is:
(x - 1)(x - 2)(x - 3)
Method 2: Rational Root Theorem
The Rational Root Theorem provides a more systematic approach to finding potential rational roots. It states that if a polynomial has a rational root p/q (where p and q are integers and q ≠ 0), then 'p' must be a factor of the constant term and 'q' must be a factor of the leading coefficient.
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Identify Potential Roots: In our polynomial x³ - 6x² + 11x - 6:
- The constant term is -6. Its factors are ±1, ±2, ±3, ±6.
- The leading coefficient is 1. Its factors are ±1.
Therefore, the potential rational roots are ±1, ±2, ±3, ±6.
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Test the Potential Roots: We test these values in the polynomial. As we discovered in the trial and error method, x = 1 is a root.
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Proceed with Polynomial Division: Just like in Method 1, divide the cubic polynomial by (x - 1) using polynomial long division or synthetic division to obtain the quadratic quotient x² - 5x + 6.
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Factor the Quadratic: Factor the quadratic x² - 5x + 6 as shown in Method 1, resulting in (x - 2)(x - 3).
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Final Factored Form: The completely factored form remains: (x - 1)(x - 2)(x - 3)
Conclusion: Mastering Cubic Factorization
Both the trial and error method and the rational root theorem are effective ways to factorize cubic polynomials. The trial and error method is quicker if you find a root early, while the rational root theorem offers a more structured approach, particularly helpful when dealing with more complex polynomials. Understanding these methods gives you the tools to confidently tackle similar cubic factorization problems. Remember to always check your work by expanding the factored form to ensure it matches the original polynomial!
