Synthetic Division in Factoring

Date: 04/18/2002 at 01:25:13
From: Brandon
Subject: Factoring completely

Factor completely: 2x^3 + 17x^2 + 58x + 25.

I don't know what to do or where to begin.

Date: 04/18/2002 at 09:01:23
From: Doctor Rick
Subject: Re: Factoring completely

Hi, Brandon.

This is not an easy type of problem to solve! The approach I would 
take is to use synthetic division to search for one root. If you are 
not familiar with this method, see the following answer in our Dr. 
Math Archives:

  Synthetic Division
  http://mathforum.org/library/drmath/view/53056.html 

Synthetic division can be viewed in two ways: as a way to evaluate a 
polynomial in x at a particular value x = a, or as a way to divide 
the polynomial by (x-a). Use this method to test possible roots. 

If the cubic has any rational roots (and if it doesn't, you won't be 
able to find an exact factorization by this method), they must be of 
the form of plus or minus a factor of 25 (the constant term) divided 
by a factor of 2 (the coefficient of the highest power). Thus the 
numerator can be 1, 5, or 25, and the denominator can be 1 or 2, and 
the sign can be positive or negative.

Synthetic division has a property that is very useful in searching 
for roots: If all the numbers below the line (in the notation used in 
the archived explanation) are positive, then you know that there are 
no roots greater than the number you tested. On the other hand, if 
the numbers below the line alternate positive and negative, then 
there are no roots less than the number you tested.

Thus, I suggest that you start by testing x = 0. This will tell you 
right away whether the polynomial may have negative roots, positive 
roots, or both. Continue in the same way (something like a "binary 
search"): each time you choose a possible root to test, choose one 
that is in the middle of the range of possible roots that you haven't 
already eliminated. It may turn out that this one test eliminates all 
possible roots less than the test number, or all possibilities greater 
than the test number, thus giving you only half as many possibilities 
to test.

Once you find a value of x for which the polynomial evaluates to zero 
(in other words, a factor (x-a) of the polynomial), you have also 
found the other factor: just use the numbers below the line of your 
synthetic division as the coefficients of a polynomial. In your case, 
the resulting polynomial will be a quadratic. Then your task is to 
factor this quadratic, if it can be done with real coefficients. (Or, 
if you have learned about polynomials with complex coefficients, you 
can always factor the quadratic.) You can use familiar methods for 
factoring a quadratic, or you can use the sure-fire method of setting 
the quadratic equal to zero and finding roots with the quadratic 
formula.

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