The polynomial \( f(x) = x^3 - 6x^2 + 11x - 6 \) can be analyzed using Vieta's formulas. For a cubic polynomial of the form \( ax^3 + bx^2 + cx + d \), the sum of the roots is given by \( -b/a \). Here, \( a = 1 \) and \( b = -6 \). Therefore, the sum of the roots is \( -(-6)/1 = 6 \).

["Analyzing the Cubic Polynomial ( f(x) = x^3 - 6x^2 + 11x - 6 ) with Vieta’s Formulas", "Understanding the roots of a polynomial is essential in algebra, and for cubic equations, Vieta’s formulas provide powerful insight without requiring explicit root computation. The polynomial ( f(x) = x^3 - 6x^2 + 11x - 6 ) is a classic example where Vieta’s relationships bring clarity to the structure and behavior of its roots.", "### The Structure of a Cubic Polynomial", "A general cubic polynomial takes the form:", "[\nf(x) = ax^3 + bx^2 + cx + d\n]", "For this polynomial, Vieta’s formulas relate the coefficients ( a, b, c, d ) to the sums and products of the roots ( r_1, r_2, r_3 ):", "- Sum of the roots: ( r_1 + r_2 + r_3 = -\frac{b}{a} )\n- Sum of products of roots two at a time: ( r_1r_2 + r_2r_3 + r_3r_1 = \frac{c}{a} )\n- Product of the roots: ( r_1 r_2 r_3 = -\frac{d}{a} )", "### Applying Vieta’s Formulas to ( f(x) = x^3 - 6x^2 + 11x - 6 )", "In our case:\n- ( a = 1 )\n- ( b = -6 )\n- ( c = 11 )\n- ( d = -6 )", "1. Sum of the roots:", "[\nr_1 + r_2 + r_3 = -\frac{b}{a} = -\frac{-6}{1} = 6\n]", "This tells us directly that the three roots sum to 6, a key piece of information for solving or interpreting the polynomial.", "2. Sum of products of roots two at a time:", "[\nr_1r_2 + r_2r_3 + r_3r_1 = \frac{c}{a} = \frac{11}{1} = 11\n]", "3. Product of the roots:", "[\nr_1 r_2 r_3 = -\frac{d}{a} = -\frac{-6}{1} = 6\n]", "These relationships not only confirm the polynomial’s internal consistency but also enable efficient factorization and root finding.", "### Factoring the Polynomial", "Given the sum of roots is 6 and the product is 6, we test integer factors of 6 (since the constant term is small and the leading coefficient is 1). Trying rational root candidates such as ( \pm1, \pm2, \pm3, \pm6 ), we evaluate:", "[\nf(1) = 1 - 6 + 11 - 6 = 0 \quad \Rightarrow \quad x = 1 \ ext{ is a root}\n]", "Using synthetic division or polynomial division, we factor out ( (x - 1) ):", "[\nf(x) = (x - 1)(x^2 - 5x + 6)\n]", "Factoring the quadratic:", "[\nx^2 - 5x + 6 = (x - 2)(x - 3)\n]", "Thus,", "[\nf(x) = (x - 1)(x - 2)(x - 3)\n]", "The roots are ( x = 1, 2, 3 ), which add to ( 1 + 2 + 3 = 6 ), confirming Vieta’s result.", "### Why Vieta’s Formulas Matter", "While factoring reveals the roots explicitly here, Vieta’s formulas remain invaluable for:", "- Confirming consistency of roots without factoring\n- These algebraic relationships aid in deriving equations and checking solutions\n- Supporting advanced techniques in polynomial interpolation, symmetry analysis, and equation transformation", "### Conclusion", "The cubic polynomial ( f(x) = x^3 - 6x^2 + 11x - 6 ) elegantly illustrates how Vieta’s formulas reveal deep structural properties. With ( a = 1 ) and ( b = -6 ), the sum of the roots simplifies neatly to ( 6 )—a testament to the power and simplicity of Vieta’s relationships. Whether solving equations, analyzing symmetry, or teaching core algebra concepts, these formulas remain indispensable.", "---", "Keywords: ( f(x) = x^3 - 6x^2 + 11x - 6 ), Vieta’s formulas, cubic polynomial, sum of roots, algebraic relationships, factoring, polynomial roots, algebra education.\nMeta Description: Analyze the cubic polynomial ( f(x) = x^3 - 6x^2 + 11x - 6 ) using Vieta’s formulas. Learn how the sum of roots equals ( 6 = -\frac{b}{a} ) and discover the roots ( 1, 2, 3 ) through elegant algebraic reasoning."]