The function \( f(x) = x^3 - 6x^2 + 11x - 6 \) has roots \( r_1, r_2, r_3 \). Find \( r_1 + r_2 + r_3 \).
["Understanding the Sum of Roots for the Cubic Polynomial ( f(x) = x^3 - 6x^2 + 11x - 6 )", "The function ( f(x) = x^3 - 6x^2 + 11x - 6 ) is a cubic polynomial, and like all cubic equations of the form ( ax^3 + bx^2 + cx + d ), it has three roots—real or complex—denoted ( r_1, r_2, r_3 ). One of the most fundamental properties of polynomial roots is captured by Vieta’s formulas, which relate the coefficients of the polynomial directly to sums and products of its roots. In particular, Vieta’s formulas allow us to compute the sum of the roots without explicitly solving for each one.", "### The Sum of Roots for a Cubic Equation", "For a cubic polynomial in the standard form:\n[\nf(x) = ax^3 + bx^2 + cx + d\n]\nthe sum of the roots ( r_1 + r_2 + r_3 ) is given by:\n[\nr_1 + r_2 + r_3 = -\frac{b}{a}\n]\nThis relationship arises from factoring the polynomial as ( (x - r_1)(x - r_2)(x - r_3) ) and comparing coefficients.", "### Applying Vieta’s Formula to ( f(x) = x^3 - 6x^2 + 11x - 6 )", "In our given function:\n- ( a = 1 ) (coefficient of ( x^3 ))\n- ( b = -6 ) (coefficient of ( x^2 ))", "Using Vieta’s formula:\n[\nr_1 + r_2 + r_3 = -\frac{-6}{1} = 6\n]", "Thus, the sum of the roots of the cubic equation is 6.", "### Verifying the Roots (Optional Insight)", "To confirm this result, we can factor the polynomial. Testing rational roots (using the Rational Root Theorem), we find that ( x = 1 ) is a root since:\n[\nf(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0\n]", "Divide ( f(x) ) by ( (x - 1) ) using polynomial division or synthetic division:\n[\nf(x) = (x - 1)(x^2 - 5x + 6)\n]\nFactoring the quadratic:\n[\nx^2 - 5x + 6 = (x - 2)(x - 3)\n]\nSo the full factorization is:\n[\nf(x) = (x - 1)(x - 2)(x - 3)\n]\nThe roots are ( r_1 = 1 ), ( r_2 = 2 ), and ( r_3 = 3 ). Their sum:\n[\n1 + 2 + 3 = 6\n]\nThis matches the result obtained via Vieta’s formula, validating both the algebraic principle and the computation.", "### Conclusion", "Finding the sum of the roots for a cubic polynomial doesn’t require solving for each root individually. Thanks to Vieta’s formulas, we can quickly compute ( r_1 + r_2 + r_3 = -\frac{b}{a} = 6 ) for any cubic equation. In this case, the roots are 1, 2, and 3—conveniently summing to 6—demonstrating how theoretical polynomial properties simplify analytical tasks.", "Whether for academic study, engineering modeling, or data analysis, understanding root relationships empowers more efficient problem-solving. When faced with a cubic polynomial, remember: the sum of your roots is simply ( -(\ ext{coefficient of } x^2) \over (\ ext{coefficient of } x^3) ).", "Key takeaway: For ( f(x) = x^3 - 6x^2 + 11x - 6 ), the sum of roots is ( \boxed{6} )."]
