#### $5,634.13Question: Let $ f(x) $ be a polynomial such that $ f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1 $. Find the sum of the reciprocals of the roots of $ f(x) $.
["Understanding the Sum of Reciprocals of the Roots: A Polynomial Insight", "When working with polynomials, especially those involving roots, one powerful technique is using Vieta’s formulas. These formulas relate the coefficients of a polynomial directly to symmetric sums of its roots — a very efficient way to compute expressions like the sum of the reciprocals of the roots without explicitly solving for each root.", "Given the polynomial:", "$$\nf(x) = x^4 - 6x^3 + 11x^2 - 6x + 1\n$$", "We are asked to find the sum of the reciprocals of the roots.", "---", "### Step 1: Identify the polynomial’s roots and their properties", "Let the roots of $ f(x) $ be $ r_1, r_2, r_3, r_4 $. The sum of the reciprocals is:", "$$\n\sum_{i=1}^{4} \frac{1}{r_i} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} = \frac{r_2 r_3 r_4 + r_1 r_3 r_4 + r_1 r_2 r_4 + r_1 r_2 r_3}{r_1 r_2 r_3 r_4}\n$$", "From polynomial theory:\n- The product of all roots $ r_1 r_2 r_3 r_4 = (-1)^4 \cdot \frac{\ ext{constant term}}{\ ext{leading coefficient}} = \frac{1}{1} = 1 $\n- The sum of products of roots taken three at a time is equal to $ (-1)^3 \cdot \frac{\ ext{coefficient of } x}{\ ext{leading coefficient}} = -(-6)/1 = 6 $", "Hence:", "$$\n\sum_{i=1}^{4} \frac{1}{r_i} = \frac{6}{1} = 6\n$$", "---", "### Step 2: Efficiently confirm using transformation", "Note that the polynomial $ f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1 $ has palindromic coefficients (coefficients read the same forward and backward). This symmetry implies that if $ r $ is a root, then $ \frac{1}{r} $ is also a root — the roots come in reciprocal pairs.", "Indeed, observe that:", "$$\nf\left(\frac{1}{x}\right) \cdot x^4 = 1 - 6x + 11x^2 - 6x^3 + x^4 = f(x)\n$$", "This confirms the reciprocal symmetry: polynomials satisfying $ x^n f(1/x) = f(x) $ have reciprocal roots. Therefore, sum of reciprocals is the same as sum of roots of $ f(x) $ scaled by leading coefficient over constant term — consistent with earlier Vieta use.", "---", "### Step 3: Final answer", "Thus, the sum of the reciprocals of the roots of $ f(x) $ is:", "$$\n\boxed{6}\n$$", "This elegant result showcases how Vieta’s formulas and polynomial symmetry simplify complex root-based computations — a key skill in algebra and higher mathematics.", "---", "Keywords: polynomial roots, sum of reciprocals, Vieta’s formulas, $f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1$, reciprocal sum, symmetric polynomials, algebra.", "Meta Description:\nFind the sum of reciprocals of the roots for $ f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1 $. Using Vieta’s formulas, the result is 6 due to reciprocal symmetry."]
