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Unit 1
Section 1: Factoring Part I
When you factor a polynomial, you simply undo polynomial multiplication (see Review
section 4).
Common Factors
The easiest factoring to do is to pull out all of the factors that are common to all of the
terms in the expression. This is the reverse of distributing one term over a bunch of terms.
Example 1: Factor the following
2x2 − 8x
Both of the terms in this expression have 2x as a factor:
2x2 − 8x = 2x · x − 2x · 4
We can pull the 2x out:
2x2 − 8x = 2x · x − 2x · 4
= 2x(x − 4)
Re-distribute the 2x to see that this factorization is correct.
Taking out common factors is the simplest factoring technique, but is easily overlooked. It
should always be done before any of the other techniques.
Special Formulas
There are many factoring formulas that you can memorize. The most important one is the
difference of squares: a2 − b2 = (a + b)(a − b).
Example 2: Factor the following completely
x2 − 9
Since this is a difference of squares, we can use the formula to get:
x2 − 9 = (x + 3)(x − 3)
To check that you’ve factored correctly, you can perform the multiplication and make
sure you end up back where you started. Two other formulas that are used often are the sum of cubes and the difference of cubes:
a3 + b3 = (a + b)(a2 − ab + b2) and a3 − b3 = (a − b)(a2 + ab + b2). These two formulas are
very similar, so it is easy to confuse the two. Take the time to check your answers using
multiplication.
Example 3: Factor the following
x3 + 8 = (x + 2)(x2 − 2x + 4)
27x3 − 64 = (3x)3 − 43
= (3x − 4)(9x2 + 12x + 16)
x4 − 25 = (x2)2 − 52
= (x2 + 5)(x2 − 5)
√
√
= (x2 + 5)(x + 5)(x − 5)
Here we had to use the difference of squares twice.
Don’t forget to multiply and check.
These special formulas can be helpful, and the difference of squares is important enough that
you ought to recognize it immediately. But most polynomials that you will need to factor
will not fit any of these formulas. You’ll need to use your head when you are faced with
a factoring problem. In order to get good at factoring polynomials, you must be fluent in
polynomial multiplication, and you must practice, practice, practice.
General Quadratics
The most common factoring problem is factoring a quadratic. If a quadratic is factorable
(not all of them are), then it will factor into two linear factors. The trick is to figure out
what those factors are.
Example 4: Factor the following
x2 + 5x + 6
We want to write this as (x±?)(x±??). Since the constant term is 6, we need
? times ?? to equal 6. The factors of 6 are ±1, ±2, ±3, and ±6. We need to
select the two numbers from this list whose product is 6 and whose sum is 5 (the
x-coefficient). So the constants in the linear factors are 2 and 3:
x2 + 5x + 6 = (x + 3)(x + 2)
Of course, we’ll need to multiply (x + 3)(x + 2) to make sure that this is correct.
If it isn’t (sometimes it won’t be), then we try again. Factoring is a difficult skill to master. You might feel as though you are just guessing
when you try to factor expressions – because you are guessing. You’ll want to make an
educated guess. Since it will be a guess, it is very important that you check your factoring
by multiplying. The more you practice, the better your guesses will be and, eventually, they
won’t be guesses at all, but the correct answer.
Example 5: Factor the following
x2 − 10x + 21
The factors of 21 are ±1, ±3, ±7, ±21. Since −3+(−7) = −1 and −3·(−7) = 21,
these are the constant terms in the linear factors. So
x2 − 10x + 21 = (x − 3)(x − 7)
Multiply (x − 3)(x − 7) to make you get x2 − 10x + 21 so that you’ll know the
factorization is correct. If the coefficient on the squared term is not one, then factoring becomes even more difficult.
In this case, you might want to try a bunch of different combinations until you see for yourself
where all of the coefficients in the quadratic come from.
Example 6: Factor the following
3x2 + x − 2
Since the constant term is −2 in this case, the only possibilities for the constant
terms in the linear factors are ±1 and ±2. Some of the possible factorizations
for the quadratic are:
(3x + 1)(x − 2),
(3x + 2)(x − 1),
(3x − 2)(x + 1),
(3x − 1)(x + 2)
If you multiply each of these out, you will see that (3x − 2)(x + 1) is the answer.
In some cases there will be more possibilities than it is feasible to list and check, so always
think about where the numbers come from when you factor.
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