SOLUTION: Find all the integral values for k for which the trinomial can be factored? z^2+kx+12 r^2+kr+20

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Question 20121: Find all the integral values for k for which the trinomial can be factored?
z^2+kx+12
r^2+kr+20
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Try this: For x%5E2+%2B+kx+%2B+12
What are the integral factors of +12?
%2B12+=+%281%29%2812%29
%2B12+=+%282%29%286%29
%2B12+=+%283%29%284%29
%2B12+=+%28-1%29%28-12%29
%2B12+=+%28-2%29%28-6%29
%2B12+=+%28-3%29%28-4%29
Now find the sum of each pair of factors:
1 + 12 = 13
2 + 6 = 8
3 + 4 = 7
(-1)+(-12) = -13
(-2)+(-6) = -8
(-3)+(-4) = -7
k can be: 13, 8, 7, -13, -8, or -7
Check:
%28x%2B1%29%28x%2B12%29+=+x%5E2%2B13x%2B12
%28x%2B2%29%28x%2B6%29+=+x%5E2%2B8x%2B12
%28x%2B3%29%28x%2B4%29+=+x%5E2%2B7x%2B12
%28x-1%29%28x-12%29+=+x%5E2-13x%2B12
%28x-2%29%28x-6%29+=+x%5E2-8x%2B12
%28x-3%29%28x-4%29+=+x%5E2-7x%2B12
You can use the same procedure for r%5E2%2Bkr%2B20