SOLUTION: Given that x^2+px+q and 3x^2+q have a common factor x-b, where p,q and b are non-zero, show that 3p^2+4q=0. b^2+pb+q=3b^2+q p=2 how should i continue this? i found =2

Algebra ->  Test -> SOLUTION: Given that x^2+px+q and 3x^2+q have a common factor x-b, where p,q and b are non-zero, show that 3p^2+4q=0. b^2+pb+q=3b^2+q p=2 how should i continue this? i found =2      Log On


   

Question 1035467: Given that x^2+px+q and 3x^2+q have a common factor x-b, where p,q and b are non-zero, show that 3p^2+4q=0.
b^2+pb+q=3b^2+q
p=2

how should i continue this? i found =2
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
This is continuation of https://www.algebra.com/algebra/homework/playground/test.faq.question.1035463.html

Recheck the process that I described and identify the TWO equations saying that each remainder expression is equal to zero.

system%28pb%2Bb%5E2%2Bq=0%2C3b%5E2%2Bq=0%29, which result from the division operations. You would now know p=2, so substituting will give you the system as two equations in two unknown variables of b and q. Solve this system.