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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 \((3x - 1)(3x + 1)(9x^{2} + 1)\).

Step by step solution

01

Identify the square roots

Firstly, one must note that \(81x^{4}\) is a perfect square since \(81 = 9^{2}\) and \(x^{4} = (x^{2})^{2}\). Similarly, \(1\) is also a perfect square as \(1 = 1^{2}\). Thus, this problem can be fit into the difference of squares formula, with \(a = 9x^{2}\) and \(b = 1\).
02

Apply the Difference of Squares Formula

Substitute \(a\) and \(b\) into the difference of squares formula which is \(a^{2} - b^{2} = (a-b)(a+b)\). So, the factored form of the polynomial is \((9x^{2} - 1)(9x^{2} + 1)\).
03

Factor Further If Possible

Notice that \(9x^{2} - 1\) is also a difference of squares with \(a = 3x\) and \(b = 1\), and can therefore be factored further by applying the difference of squares formula again. Doing this, we get \((3x - 1)(3x + 1)\). Therefore, the complete factored form of the problem is \((3x - 1)(3x + 1)(9x^{2} + 1)\).

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