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Home / constant: \(b + B = 5 \Rightarrow 1 + B = 5 \Rightarrow B = 4\)

constant: \(b + B = 5 \Rightarrow 1 + B = 5 \Rightarrow B = 4\)

February 22, 2026

$constant: \(b + B = 5 \Rightarrow 1 + B = 5 \Rightarrow B = 4\)$

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Match:
\(x^2\): \(b - a + 1 = -2 \Rightarrow b - 4 + 1 = -2 \Rightarrow b = 1\)
\(x\): \(a - b + A = 1 \Rightarrow 4 - 1 + A = 1 \Rightarrow A = -2\)
So remainder is \(-2x + 4\)? But earlier substitution gave \(3\omega + 4\), contradiction.
Wait â recalculate at \(\omega\):
$\omega^2 = \omega - 1$

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