Factoring an Equation

Date: 11/13/95 at 14:43:6
From: Anonymous
Subject: factor polynomials

The problem is to factor 7y^3 + 14y^2 - 7y.  I factor out the 7y:  

  7y(y^2 + 7y - 1)

I get stuck there.  Can I factor any further?

Date: 11/13/95 at 15:9:41
From: Doctors Ethan and Ian
Subject: Re: factor polynomials

Hi,

You were right to factor out the 7y, but you may have rushed through it a 
little too quickly:

    7y^3 + 14y^2 - 7y
    -----------------
           7y

    7y^3   14y^2   7y
  = ---- + ----- - --
     7y      7y    7y


  =  y^2 + 2y - 1


However, this isn't any easier than what you ended up with. 

You need a pair of roots whose sum is -2 and whose product is -1, and 
there aren't any integers that fit the bill.  So there are two 
possibilities:  Either the roots are real, or they're complex.  

To find out which is the case, we can look at the discriminant, 

   b^2 - 4ac            where   ax^2 + bx + c = 0

If this is negative, the roots are complex.  Otherwise, they're real.  
In this case, a = 1, b = 2, and c = -1, so

  b^2 - 4ac = 2^2 - 4(1)(-1)

            = 4 + 4

            = 8

That's positive, so the roots are real.  But what are they?  We can use 
the quadratic formula to find out:

      -b +/- sqrt(b^2 - 4ac)
  x = ----------------------     whenever  ax^2 + bx + c = 0
               2a

      -2 +/- sqrt(8)
    = --------------
           2 

      -2 +/- 2sqrt(2)
    = ---------------
            2 
 
    = -1 + sqrt(2)    OR      -1 - sqrt(2)

Now, if these are the roots of the quadratic expression, then

   0 = (y - (-1 + sqrt(2))(y - (-1 - sqrt(2))

Can that be right?  Let's multiply it out and see what we get.  (I'm going to use 'sq2' instead of 'sqrt(2)' to save space and typing.)

   0 = y^2 - [(-1 + sq2) + (-1 - sq2)]y + [(-1 + sq2) * (-1 - sq2)]

     = y^2 - [-1 + -1 + sq2 - sq2]y + [(-1)^2 - (sq2)^2]

     = y^2 - [-2]y + [-1]

     = y^2 + 2y - 1

So it ain't pretty!  But it works. 

-Doctors Ethan and Ian,  The Geometry Forum