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📅 May 5, 2026 👤 scraface

May 05, 2026

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Substitute into (6):
12a + 3(2 - 3a) = 10 \Rightarrow 12a + 6 - 9a = 10 \Rightarrow 3a = 4...
Then $ b = 2 - 3 \cdot \frac{4}{3} = 2 - 4 = -2 $
From (2): $ \frac{4}{3} - 2 + c = 3 \Rightarrow c = 3 + 2 - \frac{4}{3...
And $ d = 2 $
p(x) = \frac{4}{3}x^3 - 2x^2 + \frac{11}{3}x + 2
Thus, the remainder when $ p(x) $ is divided by $ x^4 - 1 $ is simply...
\boxed{\frac{4}{3}x^3 - 2x^2 + \frac{11}{3}x + 2}
Question: Let $ a $ and $ b $ be complex numbers such that $ a + b = 4...
We know $ a + b = 4 $, and

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