This document is a journey through pre-calculus functions told through photographs, math problems, and poetry. It discusses the challenges of learning difficult math concepts and emphasizes overcoming obstacles through perseverance. It uses examples like factoring polynomials and finding the domain of rational functions to illustrate key concepts. Poetry reflects on feeling overwhelmed but realizing the value of learning from mistakes to achieve success.

1. D.E.V. Project
A journey through pre-calculus functions told
in photographs, math problems, and “poetry”.
By Paige Sheffield

2. Impossible
When am I ever going to use this?
Challenging
I don’t have time for this.
Why did I
take this
Obstacles class?

3. It used to all make sense.

4. But then…

6. Who would've guessed
that factoring could be such a mess?
We thought this class would lead us to success,
not failing grades on our first tests
and where do we go from there?
It feels like nowhere,
but don't let go of your hope,
what you know now is not all you'll ever know.
The maximum area is 750000 ft. squared, Take a step back, just breathe
solutions don't always come easily
you're almost there
what is x? what is y?
keep going… sometimes you wonder why you even try.
it's the only path that leads to knowing. But if you keep solving,
the answers will reach you.
If you keep making mistakes,
you'll learn what to do.
Although it may be painful, although it may hurt
sometimes we have to look back to when we did the worst.
Without looking at previous mistakes,
we'll never be great.
Without looking back,
we'll keep trying to achieve success without a map.

7. Multiply the rational functions and then
find the domain of the resulting function.
Before we multiply the functions together, we must factor.

8. We’ll start by factoring 3x^3+9x^2-5x-15. For this expression, we are able to factor by
grouping. This means that we separate (3x^3+9x^2) and (5x-5), finding the greatest
common factor separately.
Notice how the sign was
changed from negative to
positive. We do this because
the groups are being subtracted
and subtracting a negative is the
same as adding a positive.

9. We’ll find the greatest common factor of (3x^3+9x^2) first.
We can look and see that the GTF is 3x^2 because both
terms are divisible by this, which leaves us with:
We then find the GTF of (5x+15), which is 5.

10. Keep in mind that each time you factor by
grouping, you should get the same expression in the
parenthesis. In this case, that value is (x+3), which is
one of the factors. We get the other factor from
what’s left after that. In this case, it’s (3x ^2-5).
We’ll come back to these factors.

11. Now we’ll factor (x^3+x^2-69x-189). To factor this, we can use long
division. First, we must find a divisor. We can find possible divisors
by diving factors of the constant by factors of the leading coefficient
(Rational Zeros Theorem). In this problem, we take the factors of 189
over the factors of 1. This gives us an idea of where to look on the table
in your calculator, but for this problem, there are several options. I
found that (x+7) works.

12. Once we have the divisor, we can begin dividing. Start by finding
what you can multiply by x to give you x^3. We get x^2, so we
write that above the bracket(first part of the quotient). Next we
multiply the x^2 by 7 and subtract (x^3+7x^2) from the first two
terms of the dividend (x^3+x^x2).
Then, we drop
the next term
down and find
what we would
need to
multiply x by to
get -6x^2. We
get -6x, and we
multiply that by
7 to get -42x.
Again, we
subtract. These
steps are
repeated until
all the terms
have been
brought down.

13. With this particular problem, we are left
with (x^2-6x-27). Now we factor this like
we would factor a quadratic normally. We
get (x+3)(x-9). We’ll come back to this
later.

14. Now we’ll factor (x^2-81). This is the
difference of squares:
a^2-b^2=(a+b)(a-b). To find the a
value, we calculate what we would
need to square to get x^2. To find the
b value, we calculate what we would
need to square to get 81.
a=x
b=9
We get (x-9)(x+9).
Lastly (as far as factoring goes for this
problem), we’ll factor (x^2+7x-
18), which can be factored like we
normally factor quadratics (by finding
two terms that multiply to equal the c
value, but equal the b value when
added together). This gives us (x-2)(x+9).

15. Now we go back to our
original functions and
multiply them in
factored form.
(x+3), (x-9), and (x+9)
divide out so we’re left
with:

16. When solving for the domain, we
look at where the denominator
equals zero. The denominator
cannot equal zero because we
cannot divide by zero. At -7 and
2, the denominator would equal
zero. Since there is no radical on
this function, the “x” value can be
negative (we can divide by
negative numbers).
Therefore, the domain is:

17. A fence is being built around the road to success. In this
problem only, the road is a dead end. Three sides will be
fenced. The people building the fence really like logarithms
and expressed the fencing they want to use as
where the “x” value
equals the perimeter in
sq. feet
What dimensions of fence will
maximize the area? x
y
x

18. Before we do anything else, we must
find the actual value of the perimeter.
We were given
sq.
feet., wher
e x=the
perimeter
To solve for x, we can use the
relationship that logs have with
exponentials.
Log x=y =
b x=b^y
This relationship tells us that
log2x=10 is the same as
2^10=x. Based on the
exponential equation, we can
find the perimeter, which is
1024 sq. feet.

19. Now we’ll set up two equations: one for
perimeter and one for area.
P=2x+y A=xy
Since we’ve already been Since “y” has been
given the perimeter, we can isolated, we can
plug that in for “P” to get now plug “1024-2x”
1024=2x+y in for y in the area
-2x -2x equation.
Now we want to get “y” We get A=x(1024-2x)
by itself, so we subtract or (1024x-2x^2)
2x from both sides. This
leaves us with y=1024-2x To find the
-1024/2(-
maximum x
2)=x
value, we use –
b/2a. X=256
Once we have the maximum x
value, we can plug it back into (1024(256)-2(256)^2)
the equation (1024x-2x^2) to Maximum area=
find the maximum area. 131072 sq. feet

20. Find the domain
We can use sum of cubes to factor (512x^3+216), but first we should look for a
greatest common factor. The greatest common factor is 8.
To find the a value, we
figure out what we would
need to cube to get 64x^3
and to find the b
Sum of cubes value, we figure out what
a^3+b^3=(a+b)(a^2-ab+b^2) we would need to cube to
get 27.

21. We get:
from using sum of cubes

22. To factor the denominator, we
can use long division. We do this
the same way we did in the first
problem.
We find the value that we can multiply
x by to get x^3, and we get x^2. We
then multiply x^2 by 8 to get 8x^2 so
we take the whole quantity of
(x^3+8x^2) and subtract it from
(x^3+9x^2). Then, we bring the 6x
down and repeat the same
steps, finding what we need to multiply
x by to get x^2. We multiply that value
(“x”, in this case) by 8 and then subtract
the whole quantity (x^2+8x) from The resulting quadratic is (x^2+x-2), which is
(x^2+6x). We get -2x when we subtract equal to (x+2)(x-1). The denominator of a
8x from 6x, and then we bring the 16 fraction cannot equal zero, so we would
down. We then find what we need to typically set up: (x+2)(x-1)(x+8)>0 to find
multiply “x” by to get -2x and then we that x cannot equal -2 ,1, or -8.
multiply 8 by the same number (-2). However, the quantity is also under a
We subtract the quantities and get radical, meaning that it cannot be
zero. negative, either.

23. The factors of the denominator are
(x+8), (x+2), and (x-1). Keeping in
mind the general shape of a
positive odd powered polynomial,
we can illustrate this graph to help
us find the domain.
Now we look to see where the
graph is positive because we
cannot take the square root of -8 -2 1
a negative. In this case, it’s
from -8 to -2 and from 1 to
infinity. However, since the * Remember that this
function is also a rational, x graph only represents
cannot equal -8, -2, or 1. This the denominator.
is because the denominator The numerator does
would be zero, and we cannot not affect the domain
divide by zero. because
, independently, it has
a domain of all real
numbers.

24. In order to continue your journey, you must factor this to
identify your license plate number (keep in mind: it’s not
supposed to be a realistic license plate number)

25. Before we can factor using long
•The first step of completing the
division, we must complete the
square is moving the constant “c”
square to get a divisor.
value to the other side of the
equation.
• We then find the perfect “c” value
by dividing the “b” value by two, then
squaring it. It’s important to
remember to add the perfect “c”
value to both sides to balance the
equation.
•We can then factor the trinomial.
The number in the binomial is going
to be half of the trinomial’s “b” value.
In this case, half of six is three. We
can use algebra to solve for x by
taking the square root of both
sides, then subtracting 3 from both
sides. We find out that x=-11 and 5.
These numbers can be used as
divisors for long division.

26. To factor this, we can do long
division just as we did in the
other problems. Once we factor
this to x^5+2x^4-
71x^3+128x^2+340x-400, we
must use another divisor to factor
it again. In this case, we can use
(x-5), which we got from
completing the square. Use the
Rational Zeros Theorem to find
other possible factors. You must
use a different divisor each time
you divide.
Rational Zeros Theorem

27. After using the Rational Zeros
Theorem, I found that (x-1) and (x-4)
are both factors.

28. Once we get a quadratic, we can factor it as we
normally would.
X^2+12x+20
We get the factors:
(x+2)(x+10)(x+11)(x-5)(x-1)(x-4)
MICHIGAN
n
5 1 4 -11 -10 -2

29. Another way to solve this is with synthetic
division. We’ll skip completing the square since
we already did that for the last method, and we’ll
move on to dividing. I’m starting with (x-5) instead of
(x+11). For synthetic division, we put
the coefficients from the original
function in the top bracket. We drop
the first number (1) down and multiply
it by five, then we add it to the second
coefficient in the top bracket (15).
This gives us 18, which we drop down
and multiply by 5 to get 90.
Therefore, we add 90 to the third
coefficient (-49) to get 41. We repeat
these steps until we get to the last
coefficient. The numbers at the
bottom are the coefficients of the
reduced function.
We get this function, and we can repeat the same steps to
divide it, again using a different divisor each time:

30. Again, we bring the first coefficient
down, multiply it by the divisor then
add it to the second coefficient. We
repeat these steps.

31. We end up getting the same quadratic that we
got when we used long division, and
therefore, the same factors (like we should).

32. Hold onto what you know,
Don’t let it go.
But clutch onto all of your mistakes,
Remember all the things that got in your
way.
Remember how impossible became
possible
Remember when rational functions
actually seemed rational
Remember when you wondered why
You even tried
And remember when you realized it was
worth it
Remember all the things that people
easily forget.

33. Overcoming obstacles
A
subject
that’s
I’ll use this all the time.
possible
The journey
continues
Thinking in
new ways
Learning from mistakes