Problem:** A function \( f(x) = x^3 - 6x^2 + 11x - 6 \) has roots \( r, s, t \). What is \( r^2 + s^2 + t^2 \)?
["SEO-Optimized Article: How to Calculate ( r^2 + s^2 + t^2 ) for the Roots of ( f(x) = x^3 - 6x^2 + 11x - 6 )", "When solving cubic equations, understanding the relationship between a polynomial’s roots and its coefficients can dramatically simplify calculations. Consider the cubic function:", "[\nf(x) = x^3 - 6x^2 + 11x - 6\n]", "This polynomial has real roots ( r, s, t ). One of the frequent challenges in algebra is computing expressions like ( r^2 + s^2 + t^2 ), especially when directly factoring or solving for roots becomes complex. This article explains how to find ( r^2 + s^2 + t^2 ) efficiently using Vieta’s formulas—without explicitly solving for each root.", "---", "### Understanding the Problem", "Given the cubic equation:", "[\nf(x) = x^3 - 6x^2 + 11x - 6\n]", "Let ( r, s, t ) be the roots. By Vieta’s formulas, we know:", "- ( r + s + t = 6 ) (sum of roots)\n- ( rs + rt + st = 11 ) (sum of products of roots two at a time)\n- ( rst = 6 ) (product of roots)", "We are tasked with finding:", "[\nr^2 + s^2 + t^2\n]", "---", "### Deriving the Expression Using Algebraic Identities", "Rather than solving for ( r, s, t ), we use a well-known algebraic identity:", "[\nr^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + rt + st)\n]", "This formula expresses the sum of squares purely in terms of symmetric sums directly available from Vieta’s.", "Substitute the known values:", "[\nr^2 + s^2 + t^2 = (6)^2 - 2(11) = 36 - 22 = 14\n]", "---", "### Why This Method Is Efficient", "Directly factoring the cubic:\n[\nx^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)\n]", "Confirms roots ( r = 1, s = 2, t = 3 ), and direct computation gives:", "[\n1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\n]", "However, this method requires exact root knowledge—and algebraic identities avoid trial and error. They work even when roots are complex or irrational, making them ideal for exam preparation and symbolic computation.", "---", "### Final Answer and Key Takeaway", "The value of ( r^2 + s^2 + t^2 ) is:", "[\n\boxed{14}\n]", "Key takeaway: For any cubic polynomial with roots ( r, s, t ), use Vieta’s formulas to compute symmetric expressions like ( r^2 + s^2 + t^2 ) efficiently—avoiding messy root-finding when possible.", "---", "Quick SEO Tips:\n- Use semantic keywords: “compute ( r^2 + s^2 + t^2 ) ro other roots,” “Vieta’s formulas,” “algebraic identities for cubic,” “sum of squares of roots cubic equation”\n- Include long-tail queries relevant to high school and college algebra students\n- Structure with headers (H2, H3) improves readability and SEO\n- Internal linking to related topics like “how to use Vieta’s formulas” boosts dwell time", "Optimized for search engines and learner intent—perfect for students mastering polynomial roots and symmetric functions."]
