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Since first two terms of \(32t^3 - 16t^2 + 8t - 4\) are divisible by \(t^2\) then the question becomes whether \(8t-4\) is divisible by \(t^2\).

(1) \(t^2 \lt 25\). Since \(t\) is a prime number then \(t=2\) or \(t=3\). If \(t=2\) then \(8t-4\) is divisible by \(t^2=4\) but if \(t=3\) then \(8t-4\) is NOT divisible by \(t^2=9\). Not sufficient.

(2) \(t^2-8t+12=0\). Solving gives \(t=2\) (\(t=6\) is not a valid solution since 6 is not a prime number). Sufficient.

I got there a different way. Can anyone clarify if my method was sound or just lucky?

First I factored out a 4 from the entire equation to yield 4*(8t^3 -4t^2 + 2t -1). Thus, the equation is a multiple of 4. If t^2 divides 4 evenly, the question can be definitively answered.

1. The stem translates to -5<t<5. Since t = prime, t = 2 or 3.

2^2 divides 4 evenly, 3^2 does not. Insuff

2. t= 2 or 6. Since 6 isn't prime t must 2 which is suff.

I believe your reasoning to be sound. Both ideas hinge on recognizing that a term that can be factored out renders the rest of the equation divisible by the term factored out. I arrived at my solution similarly.

1) -4<t<4... t must be prime so t=2 or 3. 3^2=9 can not be factored out of each term in given information. 2^2=4 can be. Insuff.

2) t= 6 or 2. T can't = 6 due to prime # restriction. 2^2=4 can be factored out of each term in original information. 4(8t^3-4t^2+2t-1). Sufficient.