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📅 May 3, 2026 👤 scraface

May 03, 2026

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📝 Recent Posts

Factor the expression \( x^4 - 16y^4 \) completely over the real numbe...
Recognize the expression as a difference of squares:
x^4 - 16y^4 = (x^2)^2 - (4y^2)^2 = (x^2 - 4y^2)(x^2 + 4y^2)
Now, note that \(x^2 - 4y^2\) is also a difference of squares:
x^2 - 4y^2 = (x - 2y)(x + 2y)
The term \(x^2 + 4y^2\) cannot be factored over the reals.
Thus, the complete factorization is:
Final answer: \(\boxed{(x - 2y)(x + 2y)(x^2 + 4y^2)}\)
A sequence of five real numbers forms an arithmetic progression, and t...
a - 6, \quad a - 3, \quad a, \quad a + 3, \quad a + 6

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