June 19th

June 19th

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