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Consider the quadratic expression 13x^2+nx-17. For certain values of n, it may be factored into a product of two linear polynomials, both of which have integer coefficients. What are all such values of n?

 Jul 18, 2021
 #1
avatar+11 
+1

welp, i just answered my own question

 

 

Since we need to transform 13x^2+nx-17 into (ax+b)(cx+d), we know that 13x^2+nx-17=axcx+axd+bcx+bd. 
ac=13, nx=axd+bcx=x(ad+bc) and -17=bd and n=ad+bc
Since 13 is prime, a or c has to be 13 or 1. 
The same thing goes for bd, since -17 is prime, b has to be 17, 1, -1, or -17 and if b is 17, d has to be -1, if b is -17, d has to be 1, if b is 1, d has to be -17 and so on.
plugging a and c in makes it either
(x+b)(13x+d) or (13x+b)(x+d) and plugging b and d in makes it either
(x+17)(13x-1) or (x-17)(13x+1) or (x+1)(13x-17) or (x-1)(13x+17).
For (x+17)(13x-1), we can expand it to 13x^2+220x-17, therefore n can equal 220
For (x-17)(13x+1), we can expand it to 13x^2-220x-17, therefore n can equal -220
For (x+1)(13x-17), we can expand it to 13x^2-4x-17, therefore n can equal -4
For (x-1)(13x+17), we can expand it to 13x^2+4x-17, therefore n can equal 4
From this, we can see that n=220, -220, -4 or 4.

 Jul 18, 2021
 #2
avatar+114112 
0

I have not looked at your question or answer, 

 

BUT it is great to see that you have answered for yourself and shared your answer with others    laugh

Melody  Jul 18, 2021
 #3
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+1

Yes, great job for figuring things out for yourself. But you can think of it in a different way: first since they both are prime numbers there common factors are 1 so for each (?x+-?) should have a 1 inside. Then after that you put in the numbers you need for 13 and 17. Multiply and your done. Try out this method too.

 Jul 20, 2021

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