# Factoring

## Graphing Rational Functions

**1.1 Factoring
**The first step in graphing a rational function is factoring polynomials. The

following is an algorithm which will generate all the factors of the form ax+b

where a and b are integers. All other real factors will require the quadratic

formula . The algorithm is described as follows for the polynomial p(x):

• List all factors of the leading coefficient and the
constant term.

• Generate a list of all possible factors ax+b where a divides the leading

coefficient and b divides the constant term.

• Test to see if each of these is a factor by plugging -b=a into p. If the

result is 0, you have found a factor.

• Do long division (or synthetic division if appropriate and you prefer)

to determine the other factor.

• Narrow your list of possible factors based on the new polynomial.

• Start over at step 3. Note that a factor may appear more than once.

**All Factors of a Number
**To find all factors of a number, test all numbers between 1 and the square

root of the number . If one of these divides evenly into the number, then you

have found two factors . As an example, consider the number 60, we will need

to test the numbers from 1 to 7 since 8

^{2}> 60.

2 divides 60 with quotient of 30

3 divides 60 with quotient of 20

4 divides 60 with quotient of 15

5 divides 60 with quotient of 12

6 divides 60 with quotient of 10

7 does not divide 60

So 60has the factors:

1; 2; 3; 4; 5; 6; 10; 12; 15; 20; 30; 60:

**All Possible Factors
**Now for every factor of the leading coefficient, a, and every factor of the

constant, b, list the factors ax + b and ax - b provided a and b do not have

a common factor . So if the polynomial were

p(x) = 6x

^{4}- 29x

^{3}+ 30x

^{2}+ 11x - 6

we use our factor list above and construct the following
factors (note that I

cycle through the q's while fixing the p).

**Test Each Factor**

The binomial ax + b is a factor of the original polynomial if and only if

p(-b/a) = 0. We work at chipping away at the polynomial until it gets

small enough that we can finish it o . For the polynomial p above:

(x - 1) p(1) = 6 - 29 + 30 + 11 - 6 ≠ 0

(x + 1) p(-1) = 6+29+30 - 11 - 6 ≠ 0

(x - 2) p(2) = 6(16)-29(8)+30(4)+11(2)-6 =0

so we do long division and

the quotient will be p_{1}(x) = 6x^{3}-17x^{2} -4x+3: Note that we can now

eliminate all of the potential factors with constant terms of 2's or 6's.

Thus we can jump to (x - 3).

(x - 3) p_{1}(3) = 6(27) - 17(9) - 4(3) + 3 = 0 so we do
long division and the

quotient will be p_{2}(x) = 6x^{2} + x - 1. Note that we can now eliminate

all factors with constant terms other than 1. Thus we could jump to

(2x - 1) as the next possible factor. However, in this case, it is easiest

just to "unfoil" it to get (3x - 1)(2x + 1).

The final result is (x - 2)(x - 3)(3x - 1)(2x + 1).

**Example**

The example is:

x^{5} - 5x^{4} - 9x^{3} + 41x^{2} + 32x - 60

The factors of 60 are given above and so we check:

(x - 1) p(1) = 1 - 5 -9 + 41+32 - 60 = 0 and so we long
divide to get

p_{1}(x) = x^{4} - 4x^{3} - 13x^{2} + 28x + 60

(x - 1) p_{1}(1) = 1 - 4 - 13 + 28 + 60 ≠ 0

(x + 1) p_{1}(-1) = 1 - 4(-1) - 13(1) + 28(-1) + 60 ≠ 0

(x - 2) p_{1}(2) = (16) - 4(8) - 13(4) + 28(2) + 60 ≠ 0

(x + 2) p_{1}(-2) = (16) - 4(-8) -13(4) + 28(-2) + 60 = 0 so
we long divide to

get p_{2}(x) = x^{3} - 6x^{2} - x + 30

(x + 2) p_{2}(-2) = (-8)-6(4)-(-2)+30 = 0 so we long divide
to get p_{3}(x) =

x^{2} - 8x + 15 which we can unfoil to (x - 3)(x - 5).

The result is (x - 1)(x + 2)(x + 2)(x - 3)(x - 5).

**1.2 Exercises
**Factor the following completely over the integers. The final answer should

consist of constants (a), binomials (ax + b) and possibly some quadratics

(ax

^{2}+ bx + c) which have irrational or imaginary roots.

1. 2x^{4} + 5x^{3} - 5x - 2

2. 9x^{4} - 24x^{3} + 22x^{2} - 8x + 1

3. x^{4} + 4x^{3} + 6x^{2} + 4x + 1

4. x^{4} - 1

5. 10x^{4} + 18x^{3} - 54x^{2} - 50x + 12

6. 3x^{3} - 32x^{2} + 83x - 42

7. 11x^{4} - 175x^{3} + 655x^{2} - 665x - 66

8. 12x^{4} - 16x^{3} - 53x^{2} - 26x + 15

**1.3 Appendix: Descarte's Rule of Signs
**We don't have time for this, but someday I will stick it here.

Draft Version: c 1998 Aaron Montgomery

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