Factoring

Date: 12/03/2001 at 21:42:09
From: Sommer Pruitt
Subject: Factoring

Hi Dr. Math, 

This problem is giving me a hard time: 6x squared + 31x + 5

I don't understand it.

Thanks for your time,
Sommer

Date: 12/04/2001 at 02:44:56
From: Doctor Jeremiah
Subject: Re: Factoring

Hi Sommer,

When you factor this you will end up with (6x+a)(x+b), because when 
you multiply those two terms together you get:

 (6x+a)(x+b)
  6x^2 + ax + 6bx + ab
  6x^2 + (a + 6b)x + ab

And the 6x^2 (6x squared) is then accounted for. Now we have two 
things to figure out: a and b.

We know that 31 = (a + 6b), and we know that 5 = ab, because those 
are the values that were in the original equation, so we have two 
equations and two unknowns:

  31 = a + 6b
   5 = ab

All we need to do is to come up with integers that work in these 
equations. Look at the second equation first. There are only really 
two possible answers for a and b. Either a = 1 and b = 5, or a = 5 and 
b = 1.

Now look at the first equation, (31 = a + 6b). The only way to get 31 
is for b to equal 5 and a to equal 1, so:

  a = 1, b = 5

Our original factors were (6x+a)(x+b), and now we know a and b so we 
can fill them in:

  (6x+a)(x+b)   <==   a = 1, b = 5
  (6x+1)(x+5)

Now all we need to do is check our answer. We can do that by 
multiplying the factors together:

  (6x+1)(x+5)
  6x(x+5) + 1(x+5)
  6x(x) + 6x(5) + x + 5
  6x^2 + 30x + x + 5
  6x^2 + 31x + 5

And since that is what we wanted, the factors really are (6x+1)(x+5).

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/