Question #24103

Factor completely:
x^4 - 6x^2 y^2 + 9y^4

Expert's answer

Let's use the perfect square formula:

x^4 - 6x²y² + 9y^4 = (x² - 3y²)^{2}

And now let's use the difference of two squares formula:

(x² - 3y²) = (x - √3y)(x + √3y).

Now factorization is complete. Therefore,

x^4 - 6x^2 y^2 + 9y^4 = (x - √3y)^{2}(x + √3y)^{2}

x^4 - 6x²y² + 9y^4 = (x² - 3y²)

And now let's use the difference of two squares formula:

(x² - 3y²) = (x - √3y)(x + √3y).

Now factorization is complete. Therefore,

x^4 - 6x^2 y^2 + 9y^4 = (x - √3y)

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