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What is the greatest integer x such that |6x^2-47x+15| is prime?

 Jan 6, 2019
 #1
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+1

x = 8

 |6x^2-47x+15|  = 23 - which is a prime

 Jan 6, 2019
 #2
avatar+701 
+2

6x^2-47x+15 can be positive or negative or 0. We can factor \(6x^2-47x+15\) into \((3x-1)(2x-15)\). For (3x-1)(2x-15) to be prime, either (3x-1) or (2x-15) must be equal to 1 or -1. There are 4 scenarios. 

 

1) 3x - 1 = 1. Simplifying, we have x = 2/3. Putting this into (2x-15), we have 4/3 - 15, which is not an integer or prime. 

 

2) 3x - 1 = -1. Simplifying, we have x = 0. Putting this into (2x-15), we have -15. Absolute value makes it 15, which is not a prime.

 

3) 2x - 15 = 1. Simplyfying, we have x = 8. Putting this into (3x-1), we have 24-1 = 23. 23 is a prime.

 

4) 2x - 15 = -1. Simplifying, we have x = 7. Putting this into (3x-1), we have 21-1 = 20. 20 is not a prime. 

 

The largest integer x such that |6x^2-47x+15| is prime is \(\boxed{8}\).

 

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