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Question

Let Δ(x)=∣ ∣ ∣x12x25x312x2+52x+2x3+3x31x+13x22∣ ∣ ∣ and ax+b be the remainder when Δ(x) is divided by x21, find 2a+b.

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Solution

Δ(x)=∣ ∣ ∣x12x25x312x2+52x+2x3+3x31x+13x22∣ ∣ ∣andax+bbetheremainderwhenΔ(x)isdividedbyx21Δ(1)=a+b=∣ ∣030744021∣ ∣=21Δ(1)=a+b=∣ ∣232702201∣ ∣=33solvingforaandbgivesa=6andb=272a+b=15

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