Factoring Polynomials
We now begin a brief discussion of the algebraic operation
called factoring. To factor an
expression means to rewrite it entirely as a product.
Thus, for example, by the time we’re done, you will be able
to determine that
3x^{ 4} y + 6x^{ 3} y  45x^{ 2} y
can be rewritten equivalently as
3x^{ 2} y (x + 5)(x  3)
Note that this second expression is a product of a monomial
factor, 3x^{ 2} y, and two binomial factors. It has just one
term, which is the product of these factors. If it had more
than one term, it would not be in factored form.
Factoring is one of the topics in algebra that many people
remember with nearly the least fondness. It seems to be a lot of
tedious work with no apparent purpose. However, as we will
demonstrate throughout these notes, the ability to factor at
least simple algebraic expressions that can be factored, can make
it possible to simplify algebraic fractions (which are even worse
to deal with than factoring, surely) and to rearrange or
manipulate formulas (which is a very important skill in many
technical applications), among other things. No matter what
people may tell you, a basic skill in factoring at least simple
algebraic expressions is an important tool in technologies that
requires some use of mathematics.
Not all algebraic expressions can be factored. Instead,
perhaps most of them cannot be factored. The method we’ll
describe in several steps in the notes to follow will let you
determine systematically and quickly whether or not an expression
can be factored, and if it can be factored, the method will
produce that factorization as part of the process.
(By the way, you should be able to verify that the first
expression above can be obtained by expanding the second
expression, using methods already described in these notes, and
so you should be able to verify that the two expressions are
mathematically equivalent. You might proceed as follows. First,
multiply the two binomials together and simplify:
(x + 5)(x  3) = (x + 5)x + (x + 5)(3)
= (x)(x + 5) + (3)(x + 5)
= (x)(x) + (x)(5) + (3)(x) + (3)(5)
= x^{ 2} + 5x  3x  15
= x^{ 2} + 2x  15
Then,
3x^{ 2} y (x + 5)(x  3) = 3x^{ 2} y (x^{ 2}
+ 2x  15)
= 3x^{ 2} y (x^{ 2} ) + 3x^{ 2} y (2x)
+ 3x^{ 2} y (15)
= 3x^{ 4} y + 6x^{ 3} y  45x^{ 2} y
