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Home » \frac4!2!1!1!0! = 12 - Aurero

\frac4!2!1!1!0! = 12 - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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2\sqrt{18 \cdot 8} = 2\sqrt{144} = 2 \times 12 = 24
Thus, the square is \(\boxed{50}\).
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: \(\boxed{(x - 2y)(x + 2y)(x^2 + 4y^2)}\)

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