# SOLUTION: How would I use the p over q method and synthetic division to factor these polynomials P(x) and then solve P(x) = 0 for each? 1. P(x) = x^4 + 11x^3 + 41x^2 + 61x + 30 2. P(x

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: How would I use the p over q method and synthetic division to factor these polynomials P(x) and then solve P(x) = 0 for each? 1. P(x) = x^4 + 11x^3 + 41x^2 + 61x + 30 2. P(x      Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Polynomials, rational expressions and equations Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Polynomials-and-rational-expressions Question 638591: How would I use the p over q method and synthetic division to factor these polynomials P(x) and then solve P(x) = 0 for each? 1. P(x) = x^4 + 11x^3 + 41x^2 + 61x + 30 2. P(x) = x^3 - 6x^2 + 3x - 10 3. P(x) = x^4 + 5x^3 + 6x^2 - 4x - 8Answer by tinbar(128)   (Show Source): You can put this solution on YOUR website!Firstly, you need to know this theorem: http://en.wikipedia.org/wiki/Rational_root_theorem If you don't understand it completely, here's what you need to know for your problem: If you have some polynomial P(x), it will certainly have some term with highest degree and some term that is just a constant. In your questions the highest degree for 1) and 3) is 4, since there is a x^4 term and no higher power. For 2) you can see the highest degree is 3 since there is a x^3 term. Now these highest powered terms have a coefficient, the number that goes in front of x^4 and x^3. In all of your questions the coefficient is 1. The constant term in 1) is 30, the constant in 2) is -10 and the constant in 3) is -8. In general the constant is the number that's by itself. Now the theorem given above basically says that if P(x) has some RATIONAL solution, then it's of the form p/q where p divides the constant term, and q divides the coefficient of the highest power. In other words, p MUST be a factor of the constant, and q MUST be a factor of the coefficient. So now let's try to check all possible roots for 1). The coefficient is 1, so 1 factors into 1*1 only, which means our solution, p/q, must have q=1. Now the constant term is 30 = 30=2*3*5. So now we have to construct all possibilities for p, p=30 or p=15 or p=10 or p=2 or p=3 or p=5 or p=6 or p=1. We consider these because any of these can divide 30 (ie, they are all factors of 30) Though we should not forget the negative versions too, since -5 also divides 30, since (-5*-6)=30. So really, for p we have, p=+-30 or p=+-15 or p=+-10 or p=+-2 or p=+-3 or p=+-5 or p=+-6 or p =+-1. Now we check them all one at a time, all combos of p/q where q=+-1 always. I'll leave it to you to check which ones work...Always start with +-1 though, it's easier, and you'll see x=-1 works, therefore (x+1) is a factor of P(x). Now divide P(x) by (x+1) and repeat the procedure.Note that the new poly you get (P(x)/(x+1) will be one degree less than P(x). So now with this 3rd degree polynomial you do the same thing as with P(x) and then once you have a 2 degree polynomial you can use the quadratic rules to solve the rest. Here's a page with a detailed solution: http://ltcconline.net/greenl/courses/103a/polynomials/FactorExample.htm