API · no tools. This is the prompt and response recorded for this attempt; a full solution may not have been returned.
Return a single integer only. The problem has been revised so that no final answer is an irrational number, a parameterized expression, a range, a proof, a drawing, a graph, a locus equation, or a region description.
Let
Find the number of integers for which is prime.
f(x) = 2x³+10x²+20x = 2x(x²+5x+10).
For f(x) to be prime, since it has factor 2x and another factor, we need the product to be prime.
f(x) = 2x(x²+5x+10).
For this to be prime, one factor must be ±1 (and the whole positive prime).
Let me consider 2x(x²+5x+10) = prime p.
Since x²+5x+10 is always... discriminant 25-40<0, so always positive.
Case: if x=0, f=0, not prime.
We need 2x(x²+5x+10) prime. The factors: 2x and (x²+5x+10).
For prime, we need one of the factors to be ±1 in absolute value such that product is prime positive.
The expression equals p. Note 2x must divide... let's think: 2x·(x²+5x+10) = p.
For positive prime, need x>0 (since x²+5x+10>0).
So x positive integer. Then 2x≥2 and x²+5x+10≥16. Product of two factors both >1 means composite unless one factor equals 1. But both ≥2 and ≥16. So never prime for x>0...
Wait, unless 2x and (x²+5x+10) share... actually product of two integers each >1 is composite. So no positive solutions.
For x negative: 2x negative, x²+5x+10 positive, product negative—not prime.
Hmm, so seemingly no primes. Let me recheck small values.
Actually wait, let me reconsider: maybe factor differently. f(x)=2x³+10x²+20x. Let me just test small integers.
x=1: 2+10+20=32. No. x=-1: -2+10-20=-12. No. x=-2: -16+40-40=-16. No. x=-3: -54+90-60=-24. x=2: 16+40+40=96.
All composite or negative. The factorization shows for x>0 product of two large factors.
So answer is 0.
\boxed{0}
Comparative Analysis (P9) (Average across runs)
Averages use reported measurements only. Missing usage is not treated as zero.