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📰 Solution: Set $ x = y = 1 $: $ k(2) = k(1) + k(1) - 2k(1 \cdot 1) = 1 + 1 - 2(1) = 0 $. Verify consistency: Try $ x = 2, y = 1 $: $ k(3) = k(2) + k(1) - 2k(2) = 0 + 1 - 0 = 1 $. Try $ x = y = 2 $: $ k(4) = k(2) + k(2) - 2k(4) $ → $ k(4) + 2k(4) = 0 + 0 $ → $ 3k(4) = 0 $ → $ k(4) = 0 $. Assume $ k(x) = 0 $ for all $ x $, but $ k(1) = 1 📰 eq 0 $. Contradiction? Wait, from $ k(2) = 0 $, check $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + k(-1) - 2k(-1) = 1 - k(-1) $. Also $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $? No: $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $. So $ k(0) = 0 $. Then $ 0 = 1 - k(-1) $ → $ k(-1) = 1 $. Then $ x = -1, y = -1 $: $ k(-2) = 2k(-1) - 2k(1) = 2(1) - 2(1) = 0 $. $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + 1 - 2(1) = 0 $, consistent. Now $ x = 2, y = -1 $: $ k(1) = k(2) + k(-1) - 2k(-2) = 0 + 1 - 0 = 1 $, matches. No contradiction. Thus $ k(2) = 0 $. Final answer: $ oxed{0} $. 📰 Question: Find the remainder when $ x^5 - 3x^3 + 2x - 1 $ is divided by $ x^2 - 2x + 1 $. 📰 Home Sweet Hell Movie 2822062 📰 Why Investors Are Obsessed Flower Foods Stock Is Breaking Records In 2024 2110091 📰 Aptenodytes Patagonicus 1664733 📰 Ready To Take Control Heres Your Certified Financial Planner Near Mefast 9967029 📰 Midiberry Is The Next Big Thingheres Why Everyone Sold On It 3879965 📰 Play Steal A Brainrotthis Gaming Hack Will Blow Your Mind Dont Miss It 7395794 📰 Pepperidge Farm Reveals A Heartbreaking Tribute To A Beloved Recipe That Changed Us All 1912633 📰 Powerphotos Mac 6220422 📰 Calculadora De Prestamos 9961086 📰 3 Hide Flawless Az Permit Test Scores 3768907 📰 Amandeuce 5096008 📰 Endicott Park Ma 7968920 📰 All Lantern Corps 1879023 📰 Shocking Waste Of Taxpayer Money How Fraud And Abuse Are Ruining Public Trust 5159372 📰 Jbg Smiths Untold Journey From Obscurity To Stardom In Just 3 Years 1442080