\omega^4 + 3\omega^3 - 2\omega^2 + \omega + 5 = a\omega + b
Related Articles
- x^4 + 3x^3 - 2x^2 + x + 5 = (x^2 - x + 1)Q(x) + ax + b
- Let \(\omega\) be a root of \(x^2 - x + 1 = 0\). Then \(\omega^2 = \omega - 1\), and \(\omega^3 = \omega(\omega - 1) = \omega^2 - \omega = (\omega - 1) - \omega = -1\). So \(\omega^3 = -1\), and \(\omega^6 = 1\).
- Now evaluate at \(x = \omega\):
- We compute each term:
- \(\omega^4 = \omega \cdot \omega^3 = \omega(-1) = -\omega\)
- \(3\omega^3 = 3(-1) = -3\)
Trending Articles
Extreme Easter Egg Hunt Backyard Survival Epic Mega Prize Locked Away Who Escapes?
Siberian Manhunt Board Game Price: Is It Worth the Icy Hype?
FNaF 4 Bedroom Minigame: Why Does This Bedroom Feel So Wrong?
Level Up Your Game with the Rarest Fable Hero Doll Ever Found
The One National Park Hidden In Plain Sight On Your Map
The One Game Where Screaming Together is Actually Deadly
Siberian Manhunt Board Game vs The Yeti: Who Will Win?
Unlock the Secret "Hand & Foot" Rules Printable That Changes Everything
Distant Melody MTG: The Forbidden Combo They Banned?
Sprintmaster Swift Silksong Trailer: 5 Shocking Twists We Missed