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\fraca^2a^2 = \boxed1 - Nelissen Grade advocaten

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Question: Let $ g(x^2 + 1) = x^4 + 2x^2 + 2 $. Find $ g(x^2 - 1) $.
g(x^2 + 1) = x^4 + 2x^2 + 2
Let’s express the right-hand side in terms of $ x^2 + 1 $.
Let $ u = x^2 + 1 \Rightarrow x^2 = u - 1 $, so:
x^4 = (x^2)^2 = (u - 1)^2 = u^2 - 2u + 1
x^4 + 2x^2 + 2 = (u^2 - 2u + 1) + 2(u - 1) + 2
u^2 - 2u + 1 + 2u - 2 + 2 = u^2 + 1
Now, substitute $ x^2 - 1 $ into $ g $:
g(x^2 - 1) = (x^2 - 1)^2 + 1 = x^4 - 2x^2 + 1 + 1 = x^4 - 2x^2 + 2
Thus, the answer is:

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