Solution: Let $ u = x^2 $, so the expression becomes $ u^2 - 13u + 36 $. Factor: $ (u - 9)(u - 4) $. Substitute back: $ (x^2 - 9)(x^2 - 4) $. Each is a difference of squares: $ (x - 3)(x + 3)(x - 2)(
["How to Simplify and Factor the Expression $ u^2 - 13u + 36 $—and Unlock Its Power Through Substitution", "When dealing with quadratic expressions, clever substitution can dramatically simplify complex problems in algebra. One such powerful solution involves letting $ u = x^2 $, transforming a seemingly challenging expression into a manageable quadratic, factoring it, and substituting back to restore the original variable. Let’s break down this elegant mathematical strategy step-by-step.", "---", "### Step 1: Start with the Original Expression\nConsider the quadratic expression:\n[\nu^2 - 13u + 36\n]", "This form appears straightforward, but recognizing and applying substitutions allows deeper insight—especially when the expression arises from real-world models or need for factorization.", "---", "### Step 2: Substitution — Let $ u = x^2 $\nBy substituting $ u = x^2 $, the expression becomes:\n[\nu^2 - 13u + 36 \ o (x^2)^2 - 13(x^2) + 36 = x^4 - 13x^2 + 36\n]\nWhile this is technically a quartic polynomial, a substitution reveals it as a quadratic in terms of $ u $, making factoring easier.", "---", "### Step 3: Factor the Quadratic in $ u $\nWe factor:\n[\nu^2 - 13u + 36\n]\nWe seek two numbers that multiply to $ 36 $ and add to $ -13 $. These numbers are $ -9 $ and $ -4 $. Hence:\n[\nu^2 - 13u + 36 = (u - 9)(u - 4)\n]", "---", "### Step 4: Substitute Back $ u = x^2 $\nReplacing $ u $ with $ x^2 $:\n[\n(x^2 - 9)(x^2 - 4)\n]\nEach factor is a difference of squares:\n- $ x^2 - 9 = (x - 3)(x + 3) $\n- $ x^2 - 4 = (x - 2)(x + 2) $", "Thus, the fully factored form is:\n[\n(x - 3)(x + 3)(x - 2)(x + 2)\n]", "---", "### Why This Technique Greatly Simplifies Problems\nFactoring using substitution breaks a higher-degree polynomial into products of linear terms, revealing roots and simplifying equation solving, graphing, and functional analysis. The difference of squares decomposition is especially powerful because it transforms nonlinear terms into recognizable univariate patterns, making computation and interpretation straightforward.", "---", "### Applications and Takeaways\nThis method exemplifies elegance in problem-solving: substituting to reduce complexity, applying foundational factoring tricks, and substituting back to restore clarity. Whether solving equations, analyzing functions, or teaching algebraic concepts, this approach empowers faster, deeper understanding.", "Key takeaway:\nLet $ u = x^2 $ when faced with $ u $-dependent quadratics — factor, decompose, substitute, and unlock streamlined solutions.", "---", "### Final Factored Form:\n[\n(x^2 - 9)(x^2 - 4) = (x - 3)(x + 3)(x - 2)(x + 2)\n]", "This complete breakdown demonstrates how strategic substitution and factoring can transform complexity into clarity—essential skills in algebra and beyond."]
