Question: Factor the following polynomial function completely. Use the Conjugate Roots Theorem, if…

Factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable.

\[ f(x) = x^6 + 4x^4 - 41x^2 + 36 \]

Answer

\[ f(x) = \]

Solution

Let’s factor the polynomial step by step. Let \( u = x^2 \). \[ f(u) = u^3 + 4u^2 - 41u + 36 & \] Find the rational roots of the cubic polynomial. Testing \( u = 1 \): \[ 1^3 + 4(1)^2 - 41(1) + 36 = 0 & \] So, \( u = 1 \) is a root. Factor out \( (u - 1) \): \[ f(u) = (u - 1)(u^2 + 5u - 36) & \] Factor the quadratic \( u^2 + 5u - 36 \): \[ u^2 + 5u - 36 = (u + 9)(u - 4) & \] Thus, the polynomial becomes: \[ f(u) = (u - 1)(u + 9)(u - 4) & \] Substitute back \( u = x^2 \): \[ f(x) = (x^2 - 1)(x^2 + 9)(x^2 - 4) & \] Factor the difference of squares: \[ x^2 - 1 = (x - 1)(x + 1) & \] \[ x^2 - 4 = (x - 2)(x + 2) & \] Putting it all together: \[ f(x) = (x - 1)(x + 1)(x - 2)(x + 2)(x^2 + 9) & \] This is the completely factored form of the polynomial.