\(x^2\): \(b - a + 1 = -2 \Rightarrow b - 4 + 1 = -2 \Rightarrow b = 1\)
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- Now add \(Ax + B\) (since divisor is quadratic, remainder is linear: \(Ax + B\)):
- x^4 + (a - 1)x^3 + (b - a + 1)x^2 + (a - b + A)x + (b + B)
- Match:
- \(x\): \(a - b + A = 1 \Rightarrow 4 - 1 + A = 1 \Rightarrow A = -2\)
- constant: \(b + B = 5 \Rightarrow 1 + B = 5 \Rightarrow B = 4\)
- So remainder is \(-2x + 4\)? But earlier substitution gave \(3\omega + 4\), contradiction.
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