ra \( 1> \) X. 8 Factor using a quadratic pattern ADW F)) Factor \( x^{4}-4 x^{2}-45 \) completely. All factors in your answer should have integer coefficients.

To factor \( x^{4}-4x^{2}-45 \), we can use substitution. Let \( y = x^2 \), transforming the equation into \( y^2 - 4y - 45 \). Now, we look for two numbers that multiply to \(-45\) and add to \(-4\). These numbers are \(-9\) and \(5\). Thus, we can factor it as \( (y - 9)(y + 5) \). Substituting back \( y = x^2 \), we get \( (x^2 - 9)(x^2 + 5) \). The first factor can be further factored as \( (x - 3)(x + 3) \), while \( (x^2 + 5) \) remains as is because it doesn’t factor over the integers. Hence, the complete factorization is \( (x - 3)(x + 3)(x^2 + 5) \). If you're looking for ways to verify your factors quickly, plug in a few values for \( x \) to see if they yield zero. This helps you confirm that \( x = 3 \) and \( x = -3 \) are indeed roots, validating that your factoring is correct!