SOLUTION: Sam is working on some polynomial factorizations in the form of x^2 + px + q , where p and q are nonzero integers and "a" and "b" are integer numbers such that x^2+px+q = (x+a)(x+b

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Sam is working on some polynomial factorizations in the form of x^2 + px + q , where p and q are nonzero integers and "a" and "b" are integer numbers such that x^2+px+q = (x+a)(x+b      Log On


   

Question 1172365: Sam is working on some polynomial factorizations in the form of x^2 + px + q , where p and q are nonzero integers and "a" and "b" are integer numbers such that x^2+px+q = (x+a)(x+b). His work is as follows:
x^2 −2x−3=(x−3)(x+1)
x^2 +5x+6=(x+2)(x+3)
x^2 −7x+10=(x−2)(x−5)
x^2 +6x+8=(x+2)(x+4)
x^2 −8x+12=(x−2)(x−6)
x^2 +9x+18=(x+3)(x+6)
He concludes that if p and q are coprime, then the factors a and b are also coprime. If p and q are not coprime, then the factors a and b are not coprime, either.
Is his conclusion correct? Explain please.
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.

So,  we have two integer numbers,  "a"  and  "b",   on one hand side,  and

        we have two other integer p and q,  on the other hand side,  such that

                p = -(a+b)   and   q = ab.


(1)    First question is:   is it true that  IF  p  and  q  are coprime,  THEN  "a"  and  "b"  are coprime ? 

     Let's prove it by CONTRADICTION.


         Let assume that "a" and "b" are not coprime.

         It means that there is such a prime number w such that "a" and "b" are multiple of w.

         Then it is OBVIOUS that w is the common divisor for (a+b) and ab.

         Hence, w is the common divisor for p and q.

         It proves, by CONTRADICTION, that if p and q are coprime, then "a" and "b" are coprime, too.


     So, the first statement is proved.


(2)    Second statement is:   IF  p  and  q  are not coprime,  THEN  "a"  and  "b"  are not coprime,  too. 

    Here the DIRECT proof works.


        If p and q are not coprime, then (a+b) and (ab)  are not coprime.


        It means that (a+b) and (ab) have common prime divisor w.


        Since w divides the product (ab), it divides at least one of the numbers "a" or "b".

            If it eventually divides both "a" and "b", then there is NOTHING to prove . . . 


        So, let assume that w divides "a". 


        Then w divides "a" and (a+b),  at the same time.


        It just implies that w divides b, too   (which is OBVIOUS).


        So, we proved that if p and q are not coprime, then "a" and "b" are not coprime, too.


    The second statement is proved.

The problem is just solved:   both statements are proved.


/\/\/\/\/\/\/\/

The post-solution note:   do not call the numbers  "a"  and  "b"  as   factors  -  in given context,  they are not.