Expert: Richard J. Raridon - 4/22/2004

Question
i still don't understand, how do you factor out those problems
1.factor 6x^3-x^2-54x-56
2.factor 2x^2-5x-3
3.factor 2x^3 + x^2-4x-1
Thank You!

Answer
Let's start with No. 2 first.  
Any quadratic equation, if it can be factored, will have factors of the form (ax-b)(cx-d).  In this case, you know that ac = 2 so you can assume a=2 and c=1 which gives you
(2x-b)(x-d).  Now bd = -3 so b = +/-3 or +/-1.  With a little trial and error, you'll find b= 1 and d = -3 which gives you (2x+1) and (x-3) for the factors.  
1. This is more complicated.   I wrote it as
f(x) = 6x^3-x^2-54x-56
Now you can put in values for x and solve for f(x).  f(x) will be zero if x is a solution.  So you try simple values like x = 1,-1,2,-2, etc.  I found that x= -2 was a solution so (x+2) is one of the factors. Then I divided x+2 into
6x^3-x^2-54x-56 and got 6x^2-13x-28.  Now you can treat it like No. 2 to get additional factors of (3x+4) and (2x-7).
3. Again, I wrote this as f(x) = 2x^3+x^2-4x-1
It doesn't have any simple factors, as least that I could find.  
f(1) = -2 and f(3/2) = 2 so there's a solution in between.
f(0) = -1 and f(-1) = 2 so there's a solution in between.  
f(-1/4) = 1/32 so that's pretty close to a solution.