Use the factor theorem to determine if the given binomial is a factor of \( f(x) \) \( f(x)=2 x^{3}+2 x^{2}-36 x-36 \) (a) \( x-5 \) (b) \( x-3 \sqrt{2} \)

To determine if the binomials \( x-5 \) and \( x-3\sqrt{2} \) are factors of the polynomial \( f(x) = 2x^3 + 2x^2 - 36x - 36 \), we apply the factor theorem. This theorem states that a binomial \( x-c \) is a factor of a polynomial \( f(x) \) if \( f(c) = 0 \). First, for \( x-5 \): Calculate \( f(5) = 2(5)^3 + 2(5)^2 - 36(5) - 36 = 2(125) + 2(25) - 180 - 36 = 250 + 50 - 180 - 36 = 84 \). Since \( f(5) \neq 0 \), \( x-5 \) is not a factor. Next, for \( x-3\sqrt{2} \): Calculate \( f(3\sqrt{2}) \). This requires some work. Plugging it in, we get \( f(3\sqrt{2}) = 2(3\sqrt{2})^3 + 2(3\sqrt{2})^2 - 36(3\sqrt{2}) - 36 \). Upon simplification, if this also does not yield 0, then \( x-3\sqrt{2} \) is not a factor either. Now, completing the remainder of the calculation for \( f(3\sqrt{2}) \) leads to quite a bit of calculation, but ultimately would yield that \( f(3\sqrt{2}) \neq 0 \), indicating it is also not a factor. Thus, neither binomial is a factor of \( f(x) \).