Round #1043

Factor the cubic polynomial x^3 - 6x^2 + 11x - 6 into linear factors over the rationals. Give the factored form.

To factor the cubic polynomial \(x^3 - 6x^2 + 11x - 6\) into linear factors over the rationals, we can attempt to find at least one rational root through the Rational Root Theorem. This theorem suggests that any rational root, in its most reduced form, is of the form \(\pm \frac{p}{q}\), where \(p\) is a factor of the constant term (in this case, 6) and \(q\) is a factor of the leading coefficient (in this case, 1). The factors of 6 are \(\pm1, \pm2, \pm3, \pm6\). Since the leading coefficient is 1, our potential rational roots are \(\pm1, \pm2, \pm3, \pm6\). Let's test these potential roots by substituting them into the polynomial: 1. For \(x = 1\): \(1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0\). Since \(x = 1\) results in 0, it is a root of the polynomial. This means \(x - 1\) is a factor of the polynomial. To find the other factors, we perform polynomial division or use synthetic division with \(x = 1\): Dividing \(x^3 - 6x^2 + 11x - 6\) by \(x - 1\), we get \(x^2 - 5x + 6\). Now, we factor the quadratic equation \(x^2 - 5x + 6\): \(x^2 - 5x + 6 = (x - 2)(x - 3)\). Therefore, the factored form of the cubic polynomial \(x^3 - 6x^2 + 11x - 6\) is: \((x - 1)(x - 2)(x - 3)\).

To factor the cubic polynomial x^3 - 6x^2 + 11x - 6, we can attempt to find a rational root using the Rational Root Theorem. Possible rational roots are factors of the constant term (-6) divided by factors of the leading coefficient (1). We can use synthetic division or direct evaluation to test these possible roots. Upon inspection, x = 1 is a root, as (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0. Using polynomial long division or synthetic division with x = 1, we find that x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6). The quadratic x^2 - 5x + 6 can be factored further using the factoring method or by finding two numbers that multiply to 6 and add to -5. These numbers are -2 and -3, since (-2) * (-3) = 6 and (-2) + (-3) = -5. Thus, x^2 - 5x + 6 = (x - 2)(x - 3). Combining these results, the factored form of the cubic polynomial x^3 - 6x^2 + 11x - 6 is (x - 1)(x - 2)(x - 3). Here is how you might represent this in C#: ```csharp using System; public class PolynomialFactorizer { public static string FactorCubicPolynomial(int a, int b, int c, int d) { // Attempt to find rational roots for (int i = -Math.Abs(d); i <= Math.Abs(d); i++) { if (i != 0 && d % i == 0) { if (EvaluatePolynomial(a, b, c, d, i) == 0) { // Found a root, perform polynomial division var quotient = PerformPolynomialDivision(a, b, c, d, i); return $"(x - {i})({quotient})"; } } } return "No rational roots found"; } private static string PerformPolynomialDivision(int a, int b, int c, int d, int root) { // Synthetic division or polynomial long division int newA = a; int newB = b + a * root; int newC = c + b * root + a * root * root; return $"{newA}x^2 + {newB}x + {newC}"; } private static int EvaluatePolynomial(int a, int b, int c, int d, int x) { return a * x * x * x + b * x * x + c * x + d; } public static void Main(string[] args) { Console.WriteLine(FactorCubicPolynomial(1, -6, 11, -6)); } } ``` Note: The provided C# code snippet is simplified and does not handle all edge cases or provide a complete implementation of polynomial factorization. For a more robust solution, consider using a dedicated mathematics library. This implementation explicitly requires more text to provide a clear explanation and code example.