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📅 May 5, 2026 👤 scraface

May 05, 2026

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Note that $ x^4 + 8x^2 + 16 = (x^2 + 4)^2 $ and $ 16y^4 - 64 = 16(y^4...
Let $ A = x^2 + 4 $, $ B = y^2 + 2 $, and $ C = y^2 - 2 $, then:
A^2 - 16BC = (A - 4\sqrt{BC})(A + 4\sqrt{BC})
But this factoring is not ideal over real numbers. Instead, observe th...
So the full expression becomes:
Now factor the remaining terms:
Letâs try factoring by grouping. Try factoring $ x^4 + 8x^2 + 16 $ a...
x^4 + 8x^2 + 16 = (x^2 + 4)^2
-16y^4 + 64 = -16(y^4 - 4) = -16(y^2 - 2)(y^2 + 2)
Letâs now expand and factor directly:

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