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

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(52790)   (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.

RELATED QUESTIONS
Sam is working on some polynomial factorizations in the form of x^2 + px + q , where p... (answered by ikleyn)

All the roots of x^2 + px + q = 0 are real, where p and q are real numbers. Prove that... (answered by ikleyn)

If x+a is a factor of the polynomial x^2 + px + q and x^2 + mx + n, then prove that a =... (answered by KMST)

If p and q are the roots of x^2+PX+q=0. Find p and... (answered by josgarithmetic,ikleyn)

NEED ANSWERS ASAP 2. Near the beginning of Lesson 5.3, a strategy for factoring... (answered by josgarithmetic)

px^2+2qx+4=0,where p and q are constant,has equal roots.express q in term of... (answered by Gogonati)

a. Explain how the development of this factoring strategy is an example of working... (answered by josgarithmetic)

The exact value of {{{cos(pi/7)+cos^2(pi/7)-2cos^3(pi/7)}}} is a rational number in the... (answered by Edwin McCravy)

if 34.0356 is expressed in the form p/q where p and q are co prime integers then what can (answered by stanbon)