The document provides step-by-step instructions for factoring polynomials, finding inverse functions, simplifying rational expressions, and graphing rational functions. It includes examples of each type of problem worked out in detail from beginning to end. The examples range from relatively simple to more complex in order to demonstrate a variety of situations that may occur.

1. Doug Bradfield

2.  The first step when factoring polynomials is to identify a greatest
common factor. For example, if you are given the function,
9x3-18x2+27x-54, then a 9 would be factored out first to create the
function 9(x3-2x2+3x-6).
• Next, the number of terms must be determined, in this case there
are four terms.
 With four terms, factor by grouping if possible.
• 9(x3-2x2+3x-6) now, break the problem into two parenthesis. We
can set the 9 aside for now if we remember to bring it back.
• (x3-2x2)+(3x-6) Then factor out anything in the two new parenthesis.
• x2(x-2)+3(x-2) x-2 must remain one factor and the other is
developed by the coefficients. This becomes 9(x2+3)(x-2)

3.  All together this problem looks like this:
 9x3-18x2+27x-54
 9(x3-2x2+3x-6)
 x2(x-2)+3(x-2)
 9(x2+3)(x-2)
 Unfortunately. Some problems are more difficult and require us to
find a rational root and use that to set up a long division problem.
 For example, x3+21x2+99x+135 is a polynomial that can’t be
grouped. One rational root must be found first. In this case x=-3 so
we use x+3 to divide out the polynomial.

4.  x+3 goes outside the division bracket and x3+21x2+99x+135 goes
inside the division bracket.
 Now the problem looks like this
 x+3 x3+21x2+99x+135
 Now, we find what times x equals x3 and put it on top of the division
bracket. In this case it is x2. put that on top.
x2
 x+3 x3+21x2+99x+135
 Then, multiply x2 by x and x2 by 3. those numbers are then
subtracted from the overall equation.

5. x2
 x+3 x3+21x2+99x+135
-(x3+3x2)
18x2
 Now, 99x must be dropped and step one must be repeated. 18x
times x equals 18x2. now 18x goes on top of the division bracket,
and is then multiplied by x and by 3. The quantity is subtracted from
the new value of 18x2+99x.
x2+18x
 x+3 x3+21x2+99x+135
-(x3+3x2)
18x2+99x
-(18x2+54x)
45x

6.  Continue these steps until all numbers have been used and there is
a remainder of zero. Here is the entire division problem.
x2+18x+45
 x+3 x3+21x2+99x+135
-(x3+3x2)
18x2+99x
-(18x2+54x)
45x+135
-(45x+135)
0
 Now, factor the function on top (x2+18x+45) directly into
parenthesis.
 15x3=45 and 15+3=18. therefore factors are 3 and 15. Don’t forget
about the x=-3 that was used to divide out.

7.  We factored x2+18x+45 into (x+3)(x+15) but after adding the x=-3
we get the final factored form of x3+21x2+99x+135 to be
(x+3)2(x+15)

8.  First, the equation must be written in y= form. For
example, f(x)=16x-7 must be written as y=16x-7.
 Next, you must solve for x. Add 7 to both sides and
divide by 16. Now we have (y+7)/16=x
 Lastly, we switch the x and y which leaves us with the
inverse. f(x)-1=(x+7)/16
 f(x)=(x+7)/16 and f(x)=16x-7 are inverses.

9. 13x+9
 If I am given the function f(x)= 2x+4 then I must write it in y= form.
This becomes y= 13x+9
2x+4
 Now you must solve for x. To do so, multiply both sides by the
quantity (2x+4). This leaves you with y(2x+4)=13x+9. After
distributing, you are left with 2xy+4y=13x+9.
 In order to get the y by itself, subtract 2xy from both sides and also
subtract 9 from both sides. Now 4y-9=13x-2xy. Now, all the x’s are
on the same side of the equation.
 From the quantity 13x-2xy an x can be factored out, leaving
x(13-2y).

10.  The equation now looks like this:
4y-9=x(13-2y)
 Divide both sides by the quantity (13-2y). Now you have
x= 4y-9
13-2y
 After switching the x and y, you are left with the inverse function
4x-9 13x+9
f-1(x)= 13-2x f(x)= 2x+4

11. 13x+9 13x+9
 f(x)= 2x+4 2x+4 ∙ y= 2x+4 ∙ 2x+4
y(2x+4)=13x+9 2xy+4y=13x+9 4y-9=x(13+2y)
-2xy -2xy (13+2y) (13+2y)
-9 -9
4y-9 13x+9
=x f-1(x)= 2x+4
13-2y

12.  The first step is always to factor out each expression whether you
are dealing with multiplication, addition, division, or subtraction. If I
am given the expression: 2x2+4x-16 ∙ 4x2+12x+8
x+2 x+4
 That can be factored into 2(x+4)(x-2) ∙ 4(x+2)(x+1)
x+2 x+4
 In multiplication, Any similar expressions can then be divided out,
which would leave us with (2x-4)∙(4x+4). In this particular case,
both denominators were able to divide out. After factoring out a 2,
the final expression in factored form is: 2(x-2)(2x+2)
 In the case of addition or subtraction, we would first have to find
common denominators before dividing out and in the case of
division, we multiply the first part of the expression by the reciprocal
of the second part.

13. x+5 x2+2x-15
 If we are given the expression: x2-1x-6 The first step, as
x2+3x-18
always, is to factor. This factors nicely into the expression:
x+5 (x+5)(x-3)
(x-3)(x+2) We now must flip the second part and multiply,
(x-3)(x+6)
called multiplying by the reciprocal. This leaves us with:
x+5 • (x-3)(x+6) The next step is to divide out. In this case, you
(x-3)(x+2) (x+5)(x-3)
can divide out an (x-3) and an (x+5). Now we are left with:
x+6
(x-3)(x+2) This is the final simplified rational expression.

14.  Here is all the work put together.
 x+5 x2+2x-15
x2-1x-6 x2+3x-18
x+6
x+5 (x+5)(x-3)
 (x-3)(x+2) (x-3)(x+6)
(x+2)(x-3)
x+5 (x-3)(x+6)
 (x-3)(x+2) • (x+5)(x-3)
x+5 (x-3)(x+6)
 (x-3)(x+2) • (x+5)(x-3)

15.  The first step to graphing a rational function is always to factor. If
we are given the function: x2+10x+9 we can factor the numerator and
x2+3x+2
denominator.
(x+9)(x+1)
 Now we have: (x+2)(x+1)
 Next, we need to look to see if any factors will divide out. In this
case (x+1) divides out leaving us with:
x+9
x+2
 There will now be a hole where we divided out. We then have to
set (x+1) equal to zero. This leaves us with x=-1
 Now we need to find any asymptotes and intercepts.

16.  In order to find Vertical asymptotes, we need to look for any
solutions that cause the denominator to be equal to zero. Our
equation only has one vertical asymptote at x=-2.
 Next, we would look for any horizontal asymptotes. In order to find
the horizontal asymptote, if the x’s are to the same power in the
numerator and denominator, we divide their leading coefficients. In
this case our horizontal asymptote is at y=1.
 To find any y intercepts, simply set x=0 in the equation and solve. In
this equation when x=0, it gives us 9 also known as 4.5. our y-
2
intercept can be found at (0,4.5)
 Lastly we must find the x-intercept. To do so we must find what
number, if plugged in for x would make the numerator=0. Our x-
intercept is -9 and can be found at (-9,0).

17.  The graph would look something like this:
_ _ __ _ _ _ _ _ _ _ _ _ _ _
hole
4.5
_ _ _ _ _ _ _ _ 1_ _ _ _ _ _ _ _
-9 -2 -1

18.  Now lets try one a little more difficult.
x+7
 x2+14x+24 following the steps, we must factor before we do
anything else.
x+7
(x+12)(x+2)
 This time we have nothing to divide out and are left with more
factors.
 Now in order to make the denominator 0, we need x to equal -12
and -2. Therefore we have 2 vertical asymptotes. One at x=-12 and
one at x=-2.
 In this situation there is a higher power in the denominator than the
numerator. This means our horizontal asymptote is y=0.

19.  Now we must find intercepts. To find the y-intercept we plug zero in
7 7
for x. After that, we are left with: 24 y-intercept is y= 24
 Lastly we must find the x-intercept by setting the numerator equal
to zero. The x- intercept is at x=-7. So far we have everything
except the direction of our arrows. To find this we must set up a
chart to find the direction as our asymptotes are approached.
x -12.1 -11.9 -2.1 -1.9
x+7 (-) (-) (+) (+)
(x+12)(x+2) (-)(-) (+)(-) (+)(-) (+)(+)
y -∞ ∞ -∞ ∞

20. 
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
-12
-7
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
-2
.29
The graph would look something like this:

21.  This project was something that I decided I would not stop until I believed I
have put all I could into it. I tried to use problems that were somewhat
challenging and show a small variety of possibilities that could occur in
each of the topics. As it turns out, I am very glad that I chose my due date
during break because of the amount of time and effort that it took to
complete and read back through checking my work. As a whole, I have
gotten much better at the topics that I covered after going back through
notes, and researching, and in some cases just figuring it out. A couple of
these topics were not my strongest during the year, but I now feel that they
are definitely my best four topics. Although this was a project that for the
longest time was nothing but confusing, I strongly recommend that you
continue to do it due to the amount of knowledge I have gained in the
process. The most difficult part for me was to come up with numbers that
would work to factor by dividing. Eventually I developed a strategy but
other than making symbols in power point, it was the most time consuming.
All in all, I would do this project again.