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Chapter 0: Problem 48

Factor the difference of two squares. $$81 x^{4}-1$$

Short Answer

Expert verified
The factored form of \(81x^4 - 1\) is \((3x - 1)(3x + 1)(3x 鈥� i)(3x + i)\)

Step by step solution

01

Identify a^2 and b^2

First, identify what \(a^2\) and \(b^2\) are. In this case, \(a^2 = 81x^4\) and \(b^2 = 1\). Therefore, \(a = 9x^2\) and \(b = 1\).
02

Use the difference of squares formula

Now, plug \(a\) and \(b\) into the formula \(a^2 - b^2 = (a - b)(a + b)\). Doing so, we get: \(81x^4 - 1 = (9x^2 - 1)(9x^2 + 1)\). But, both factors are still a difference of squares and can be factored further.
03

Further factorization

Using the same formula on the two obtained factors from the previous step, we get: \(9x^2 - 1 = (3x - 1)(3x + 1)\) and \(9x^2 + 1 = (3x 鈥� i)(3x + i)\) where \(i\) is the imaginary unit. Therefore, the fully factored form of the expression is: \(81x^4 - 1 = (3x - 1)(3x + 1)(3x 鈥� i)(3x + i)\).

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