Problem 69 - Algebra and Trigonometry

Question

Factor completely, or state that the polynomial is prime. $$2 x^{4}-162$$

Step-by-Step Solution

Answer

Hence, the fully factored form of the given polynomial \(2 x^{4}-162\) is \((\sqrt[2]{2}x^2 + 9)(\sqrt[2]{2}x^2 - 9)\).

Step 1: Identifying the Form of the Polynomial

Identify the form of the polynomial. In this case, it is a difference of two squares, which can be factored using the formula \(a^2 - b^2 = (a + b)(a - b)\). Here, \(2x^4\) can be seen as \((\sqrt[2]{2}x^2)^2\) and \(162\) can be written as \((9)^2\), hence our \(a = \sqrt[2]{2}x^2\) and \(b = 9\)

Step 2: Applying the Difference of Squares Formula

Apply the difference of squares formula to the polynomial. Following the formula \(a^2 - b^2 = (a + b)(a - b)\), we get \((\sqrt[2]{2}x^2 + 9)(\sqrt[2]{2}x^2 - 9)\)

Step 3: Simplifying the Result

Simplify the two resulted polynomials. But in this case, these two polynomials cannot be factored further so the final answer would be \((\sqrt[2]{2}x^2 + 9)(\sqrt[2]{2}x^2 - 9)\)