Question 1208350: Show that x^2 + 9 is prime.
I will first list the pairs of integers whose product is 9.
1, 9
-1, -9
3, 3
-3, -3
2. Compute their sums.
1 + 9 = 10
-1 + (-9) = -10
3 + 3 = 6
-3 + (-3) = -6
I can also say x^2 + 9 = x^2 + 0x + 9 and none of the sums added above equal 0. This leads me to conclude that x^2 + 9 is prime.
1. Do you agree?
2. How is this done for a trinomial?
Sample:
Show that x^1 + x + 1 is prime.
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
You clearly have not stated the problem correctly; or else the problem is faulty.
If x is any multiple of 3, then x^2+9 is not prime.
If x is any odd integer, then x^2 is odd, so x^2+9 is even, so it is not prime.
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To the student, who suggests in his "thank you" note that I didn't know what to do with his question....
It is YOU who does not know what to do. You posted a problem without the required context.
Without any context, when you ask "is x^2+9 prime", the usual interpretation is that you are talking about prime numbers.
If you had posted your problem to ask whether "x^2+9 is a prime polynomial", then you would have saved both of us some wasted time.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
You have the correct reasoning for problem 1.
x^2+9 cannot be factored into the form (x+a)(x+b) where a,b are integers.
A polynomial is considered prime if we cannot factor it any further.
The x^1 + x + 1 you wrote out should be something like x^2 + x + 1
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