Three Factoring Techniques for Quadratic Polynomials

Date: 02/21/2008 at 17:48:06
From: Sarah
Subject: factor the following: 3(40x^3+31x^2+6x)

How would you factor 3(40x^3+31x^2+6x)?  There is an exponent to the
3rd power, and the numbers are too large and can not be reduced any more.

The original question was 120x^3+93x^2+18x, but I reduced that.  I 
also know that the parenthesis should look like this...
3(_x__)(_x__)(_x__), but what goes into the underlined areas?????

Date: 02/21/2008 at 23:17:53
From: Doctor Greenie
Subject: Re: factor the following: 3(40x^3+31x^2+6x)

Hi, Sarah --

Let's look at the original problem:

  120x^3 + 93x^2 + 18x

You saw the common factor of 3; but there is also a common factor of 
x.  So this original expression can be factored to

  3x(40x^2 + 31x + 6)

So if the quadratic expression factors, the final factored form will 
be of the form

  3x(_x + _)(_x + _)

Now we have to deal with that quadratic expression with the large 
coefficients.  There are several methods that can be used to do 
this, and there are often many small variations of a particular 
method.  I will demonstrate three methods that I have found easy to 
use (IF you understand how to apply them!!).

We need to factor the expression

  40x^2 + 31x + 6  ( = ax^2 + bx + c )

1. The "ac" method

One method commonly taught is to multiply the "a" and "c" 
coefficients together to get a "target number":

  ac = 40*6 = 240

Then our task is to find two numbers whose product is 240 and whose 
sum is the coefficient of the middle term, 31.  We find the two 
numbers to be 15 and 16.  The next step is to write the middle term 
of the original expression, 31x, using these two numbers--that is, we 
write the middle term as 15x + 16x.  When we do that, we can then 
finish the factoring by grouping.

  40x^2+31x+6
  (40x^2+16x) + (15x+6)
  8x(5x+2) + 3(5x+2)
  (8x+3)(5x+2)

Note if you try using this method that, on the third line in this 
example, you should end up with a common binomial factor if you have 
done the rest of the process correctly.

2. Organized trial and error

I learned factoring by this basic method; and I have just recently 
been shown a way to organize this method so that a lot of the 
guesswork is removed.

We have to start with a guess of some sort.  For this example, I will 
start by guessing that the leading terms of the binomial factors are 
10x and 4x.  I start the process by writing these leading coefficients 
10 and 4 like this:

   4 |
  ---------
  10 |

On the right side of the "|" separator, I write all the possible 
combinations of factors that give me the constant product of 6:

   4 | 6 1 3 2
  -------------
  10 | 1 6 2 3

This display shows that, IF the leading coefficients are 4 and 10, 
the POSSIBLE factorizations are

  (4x+6)(10x+1)
  (4x+1)(10x+6)
  (4x+3)(10x+2)
  (4x+2)(10x+3)

But every one of these possible factorizations contains a factor which 
contains a common factor.  (And note that we can see that this will be 
the case by looking at our array of numbers--we don't need to write 
out all the possible factorizations).  Since the original expression 
does not contain a common factor, none of these can be the correct 
factorization of the original expression.  So the leading coefficients 
of our factors can't be 10 and 4.

So let's try leading coefficients of 8 and 5 and go through the same 
process.

   8 | 6 1 3 2
  -------------
   5 | 1 6 2 3

We can immediately reject the possibilities which pair the "6" or "2" 
with the leading coefficient of 8, because those would produce factors 
with a common factor.  So we only need to look at the remaining 
possibilities:

   8 | 1 3
  ----------
   5 | 6 2

To get the middle term of the product from this array, we can perform 
a sort of "cross multiplication".  If we choose the "1" and "6" 
constant terms, then the coefficient of our middle term is going to be 
8(6)+5(1) = 48+5 = 53--not what we want.  But when we choose the "3" 
and "2" constant terms, we get the desired middle term:  8(2)+5(3) = 
16+15 = 31.

So we find that the factorization of our expression is

  (8x+3)(5x+2)

3. Pure magic

Well, it's not really magic--it's just hard to show why this method 
works; and it's hard to UNDERSTAND why it works.  But in the few cases 
where I have seen it used, students who understand it find it quite 
easy to use....

We start like the ac method described above--we multiply the 40*6 to 
get 240.  But this time we make a new equation with a leading 
coefficient of 1 and a constant term of 240:

  x^2 + 31x + 240

Next we factor this by usual methods.  Since the leading coefficient 
is 1, it is relatively easy to find the numbers whose product is 240 
and whose sum is 31; those numbers are again 15 and 16.  So we 
factor this new expression as

  (x+15)(x+16)

Now comes the "magic" part.  We replace each "x" in this factored form 
with "40x"--using the leading coefficient of our original expression:

  (40x+15)(40x+16)

Finally, we remove the common factor from each of these factors:

  (8x+3)(5x+2)

Abracadabra!  Poof!  Our "magic" method has given us the correct 
factorization.

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