Use the remainder theorem to determine if the given number \( c \) is a zero of the polynomial. \( q(x)=x^{3}-4 x^{2}+4 x-16 \) (a) \( c=2 i \) (b) \( c=-2 i \)
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To determine if \( c \) is a zero of the polynomial \( q(x) = x^3 - 4x^2 + 4x - 16 \), we can use the Remainder Theorem, which states that if we evaluate a polynomial at \( c \), the result must be zero for \( c \) to be a root. First, let's evaluate \( q(2i) \): \[ q(2i) = (2i)^3 - 4(2i)^2 + 4(2i) - 16 = -8i - 16 - 16 + 8i = -32. \] Since \( q(2i) \neq 0 \), \( 2i \) is not a zero of the polynomial. Now, let's evaluate \( q(-2i) \): \[ q(-2i) = (-2i)^3 - 4(-2i)^2 + 4(-2i) - 16 = 8i - 16 - 8i - 16 = -32. \] Since \( q(-2i) \neq 0 \), \( -2i \) is also not a zero of the polynomial. Thus, neither \( 2i \) nor \( -2i \) is a root of \( q(x) \).