Factoring an Equation

Date: 3/11/96 at 19:43:58
From: Anonymous
Subject: Algebra/factoring

My math problem is if you have the equation

   x2 (x squared) * (times) 2x + 3 = 0 

how do you factor it?

I have tried many ways and I have gotten completely lost so if you
could start at the beginning I would really appreciate it.  The
teacher says that it will come out to be some answer without 
decimals.

Thank You,
   Melissa Lott

Date: 3/12/96 at 10:3:56
From: Dr. Elise
Subject: Re: Algebra/factoring

Hi Melissa,

I see your dilemma.  The problem you have stated is:

x^2 * 2x + 3 = 0  Which is really:

2x^3 + 3 = 0

2x^3 = -3

x^3 = -3/2    I'm sure even your teacher will agree that this is not
              going to turn out as "some answer without decimals".  The
              last time I checked, the cube root of negative three-
              halves was an imaginary decimal.  No wonder you're 
              completely lost!

              My best guess is that something went wrong when you copied
              the problem - I'd advise you to check it out.  I'll show 
              you what I'll bet the problem was, though!

Okay.  Factoring.  Let's say that the "times" should really have been 
a "plus".

x^2 + 2x + 3 = 0

Now we go through the usual factoring process.  The coefficient of 
the x^2 term is 1, so we don't worry about it any more - both factors 
are going to look like (x plus or minus something).

The 3 is a prime number, so the only way we're going to be able to 
factor it is 1 * 3.  So we know our answer is going to look like 
either:

(x + 1)(x + 3)  or  (x - 1)( x - 3)  in order to come out with a 
positive x^2 and a positive 3.  Does either of these combinations give 
us a positive 2x?

Nope.  So we know that this particular problem isn't factorable.

Let's say the problem was:

x^2 + 2x - 3 = 0  instead.

We still end up with 1 * 3 for the 3, but it has to be either - 1 * 3 or
1 * -3 to get a negative 3.  And we know we want a positive 2x, so 
we know that the bigger number, 3, has to be the positive one 
(otherwise we'd end up with a negative 2x).  So we get

(x - 1)(x + 3)  = 0

This is the only way that I can see to factor anything close to what's 
in your problem into an answer with whole numbers.  Does this 
help?

Even if the problem was:

x^2 * (2x + 3) = 0

then we still end up with either x^2 = 0  or (2x + 3) = 0,
and our solutions would be x = 0, x = 0, (twice, because it's x^2,
so it has 2 factors) and x = - 3/2, which is a fraction.

Good luck, and let me know how it turns out.

- Dr. Elise  The Math Forum