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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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