Factoring Polynomials of Degree 2

Date: 4/8/96 at 13:39:19
From: Lorrane Chung
Subject: Factoring polynomials
Hello,

I am having a lot of difficulties factoring polynomials of the 
type x^2 + 6x +8 and 3x^2 + 10x +8.  I have exhausted all the 
methods - they don't seem to work for me.  I really need an easy 
method to factor other than the quadratic formula or for finding 2 
factors that multiply to give last term but add to give the middle 
term.  Please help.

Desperate,
Lorrane

Date: 4/16/96 at 19:33:19
From: Doctor Patrick
Subject: Re: Factoring polynomials

Hi Lorrane!  I hope we can help you.  

How did you try factoring these problems?  When I have to factor, 
I figure out the possible factors for the first and last terms and 
then play with them a little to see what might work, and what is 
definitely going to be out of the question.

In the first problem, all we need to factor is the 8 - the x^2 is 
going to have to be x*x.  This gives us two choices: 8*1, or 2*4.  
Since all of the numbers in the problem are positive, we can 
ignore the other two possibilities of -8*-1 and -2*-4.  

Working with the 8*1 first, we get (x+8)*(x+1), which doesn't give 
us the middle term we are looking for.  Since there are no other 
combinations using 8*1 (do you see why?) we can move on to the 
other possibility of 2*4.

If we try factoring the equation with these numbers we get 
(x+2)(x+4). When we multiply this back out we get x*x +2x + 4x +8 
which equals x^2+6x+8 after we combine like terms and multiply out 
the x*x.

The second problem is more complicated since we have one 
additional term (the 3 in front of the x^2).  Again, we have two 
ways to factor 8, but we also need to try multiplying the 3 by 
both of the factors to find out which combination works.  If we 
start with 8*1 as factors of 8, again we get two possibilities:
(3x+8)(x+1) or (3x+1)(x*8).  

Do you understand why there are two possibilities this time?

The first possibility would end up giving us (3x*1) + (x*8), or 
3x+8X, for the middle term.  Since this would equal 11x it is not 
the combination we are looking for.  

Likewise (3x+1)(x*8) will not work.  (Why not?)  So now have to 
move on to the other set of factors (4*2).  Why don't you try the 
remaining two possibilities on your own using the same methods we 
used above.  Good luck!

Write us back if you need more help!

-Doctor Patrick,  The Math Forum