Clean Up
The first thing we need to discuss is our old nemesis from Algebra – long division of
polynomials! Let’s review this concept 1st and then I get to show you an easy way to do
long division, called synthetic division. We’ll start with a quick review of long division
with numbers, because the same principles used there are the principles used in
polynomials.
Recall:
2 7 ⎡2 8 4
A mixed number is actually a sum 10 + 14/27
Also Recall:
Finally Recall:
To check a long division multiply the quotient & divisor and
add in the remainder.
(27 x 10) + 14 = 270 + 14 = 284 CHECK
When we do long division we are really looking to see if the highest place value will go
into the number with the same place value (degree). If it won’t (say it was 25 instead of 28),
then we would have to move to the hundreds place – this is really what we are doing in
long division of polynomials too, except that we have degree instead of place value to
consider.
Polynomial Long Division:
1.
Order the polynomial dividend & divisor
2.
Divide the 1st term of the dividend by the 1st term of the divisor
3.
Multiply step 2 by divisor (don’t forget it is a binomial!)
4.
Subtract step 3 from 1st and 2nd terms (this can be done by changing the
sign of the quotient before multiplying)
5.
Bring down next term of dividend
6.
Continue the process in steps 2-5 until the remainder is of lower degree
than the divisor binomial
7.
At the point where division stops write the remainder over the divisor and
add it to the quotient
Example:
You Try:
Divide using long division
(2x2 − 5x + 9) ÷ (x − 2)
a3 − a + 5a2 + 1
a − 1
Note: Don’t forget to order the polynomial!
Y. Butterworth
Clean Up Material
Synthetic Division, Rational Eq. Apps, Variation
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In Intermediate Algebra, we learn a new way to do polynomial division. It is just a
shortcut to long division with some restrictions. First, we can only divide by a divisor of
the 1st degree, and second we must put the divisor in x − c form so that we do not have to
do subtraction. Let’s practice the x − c form first.
Example:
Put the following into x − c form:
x + 2
Steps for Synthetic Division:
1.
Order the dividend & divisor
2.
Write the divisor in in x − c form
3.
Write the numeric coefficients of the dividend and c like so
c⎦ a b d e
4.
Bring down a
5.
Multiply c and a & put under b
6.
Add b & ca
7.
Continue the process until done
8.
Answer is of 1 degree less than the dividend. Write the answer with
variables and remember that the last number is the remainder and must
be written that number over the divisor added to the polynomial
Example:
Use synthetic division to divide
(2x2 − 5x + 9) ÷ (x − 2)
Note: This should be the same as your answer to b) above.
You Try:
Divide using synthetic division
2
a)
(a − a − 13) ÷ (a + 3)
b)
(x2 + 3x + 2) ÷ (x + 1)
Now for a theorem that links synthetic division to something that we have done before –
evaluation! The true value of this theorem does not become clear until the next class (PreCalculus) but know that it does have more value than just a shorter way of evaluating a
polynomial – which it can provide!
Remainder Theorem
If P(x) is divided by x − c then the remainder is P(c)
Example:
Y. Butterworth
Find P(2) if P(x) = 2x2 − 5x + 9
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Notice what we got? It agrees with the first synthetic division example’s remainder?
It should because that is what the Remainder Theorem states will happen. Whether
you evaluate or use synthetic division you will know if a polynomial is evenly divisible
by a particular binomial. This is useful in Pre-Calculus.
Your book asks the question a little differently than I have above – the authors ask what
the value of P(2) would be and expect you to use synthetic division to get the value (by
looking at the last term, since the remainder is the same as evaluating the polynomial at “c”).
Rational Expression Equations and Their Applications
Let’s review solving equations containing rational expressions one more time.
Solving a Rational Expression Equation
Step 1: Find zeros (restrictions) of the rational expressions
Step 2: Find the LCD of the denominators in the equation
Step 3: Multiply all terms by the LCD (this is multiply, cancel, multiply out, to clear)
Step 4: Solve appropriately (may be either a linear or a quadratic, so be careful!)
Step 5: Eliminate extraneous solutions as possible solutions to the equation and write as
a solution set.
Example:
Find all value(s) of a such that
You Try:
y
+
2y + 2
2(y − 8)
4y + 4
f(a) = -2 when f(x) = 2x − 4
x
= 2(y − 1) − 1
y + 1
Now, there are 3 types of application problems that we want to discuss: number
problems, distance problems and work problems.
Number Problems
I don’t have any more to say about these problems than I already said in previous chapter
– these are just translation problems with the added bonus of having a reciprocal included
in the translation.
Example:
Y. Butterworth
The denominator of a fraction is 6 more than its numerator. If ½ is
added to the fraction the result is 27/26.
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Synthetic Division, Rational Eq. Apps, Variation
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You Try:
Three times the reciprocal of a number equals 9 times the
reciprocal of 6. Find the number.
Work Problems
These problems involve work being done by two forces. Each force is usually said to be
able to complete the entire job in some amount of time, so we narrow everything down to
the rate at which the work can be completed in one unit of time by taking the reciprocal
of the time it takes to do the entire job. The rate that it takes both forces to complete the
job in 1 unit of time will sum to the reciprocal of the total time that it will take both
working together to actually complete the job. This total time is usually the unknown. If
the forces work together it will be a sum and if the forces oppose one another it will be a
difference.
Example:
In 2 minutes, a conveyor belt moves 300 pounds of tin
from the delivery truck to a storage area. A smaller belt
moves the same amount the same distance in 6 minutes. If both
belts are used, find how long it takes to move the cans to the
storage area.
Minutes
Part of Job in 1 Minute
Large
Small
Together
You Try:
One hose can fill a pond in 45 minutes and two hoses can fill the same
pond in 20 minutes. Find how long it takes the second hose to fill the
pond alone.
Distance Problems
Distance problems will use the rate equation, t = d/r, not once but twice. Sometimes there
will be two opposing forces at work such that one adds to the rate and one subtracts from
the rate. The final equation comes from the fact that the time to complete the task with
either of the opposing forces is the same – t1 = t2, where t1 = d1/r1 and t2 = d2/r2. The
Y. Butterworth
Clean Up Material
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easiest way to keep track of all the components is a table that tracks d, r and t for the two
opposing forces. All that needs to be written outside the table is that t1 = t2 or t1 + t2 =
Given
d
r
t
Downstream, With Given distance
Given rate + rate
GivenDistance/Given rate
Wind, Downhill,
of thing in 1st
+x
etc.
column (this is
the unknown, x)
Upstream, Against Another given
OtherGivenDistance/Given
Given rate − rate
st
Wind, Uphill, etc. distance
of thing in 1
rate − x
column (this is
the unknown, x)
t1 = t2, meaning
Given Distance
GivenRate + x
t1 + t2 = Given, meaning
Example:
Distance =
You Try:
Y. Butterworth
=
OtherGivenDistance
GivenRate − x
Given Distance +
GivenRate + x
Other Given Distance
GivenRate − x
= Given
A cyclist rode the first 20-mile portion of his workout at a constant
speed. For the 16-mile cool down portion of his workout, he
reduced his speed by 2 miles per hour. Each portion of the
workout took the same time. Find the cyclist's speed during the
first portion and find his speed during the cool down portion.
Rate •
Time
Ray starts out on a boating trip early one morning. Ray's boat can go
20 mph in still water. How far downstream can ray go if the current is
5 mph and he wishes to go down and come back in 4 hours?
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Synthetic Division, Rational Eq. Apps, Variation
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Variation
Our last topic is variation. This is simply a matter of learning to translate a few phrases
into equations and then using the translation to solve for an unknown.
Direct variation means that two numbers vary proportionally by a constant. Another way
of saying this is that one is a constant multiple of the other. The following is the
translation of y varies directly (proportionally) with x.
y = kx,
where
y and x ∈ ℜ
and
k is a constant of proportionality.
Example:
a)
When y varies directly with x, find the constant of
proportionality, k.
y = 10 and x = -2
b)
y = 7 and x = -3
The constant of proportionality is always the same for a given relationship and this can be
used to solve for unknowns, using a proportion. Since K is always equal to y/x when x
and y vary proportionally and the constants of proportionality are always equal for a
given relationship, we can set up a proportion and solve it based upon this unique
constant, K.
Example:
a)
Given that y varies directly with x, solve the following.
x = 5 and y = -2, find y when x = 20.
Inverse variation means that two numbers vary inversely by a constant. Another way of
saying this is that one is a constant multiple of the other’s reciprocal. The following is
the translation of y varies inversely (indirectly) with x.
y = k/x,
where
y and x ∈ ℜ
k is a constant
and
Example:
a)
Y. Butterworth
When y varies inversely with x, find the constant of
proportionality, K.
y = 10 and x = -2
b)
y = 7 and x = -3
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There are many things in our real world that vary directly and indirectly. For instance,
distance and time vary directly and when you consider the formula d = r • t, and time
and rate vary inversely if you look at this equation rewritten as t = d/r. There are many
other relationships that occur in our physical world that can be described like this and
many others that we can concoct to show this, but I will focus on just this one to show
some applications of the idea.
Example:
You Try:
Distance and time vary directly, when rate is constant. If
a car travels 250 miles in 3 hours, how many hours will it
take to travel 700 miles?
The intensity of a light sources varies inversely with the square of the
distance from the source. The intensity of a light is 36 foot-candles at
3 feet, find the intensity at 6 feet.
Joint variation or jointly proportional means that one variable increases or decreases as
two others increase or decrease.
y varies jointly with x&z
Example:
translates to
y = kxz
The area of a triangle varies jointly with the triangle’s
base and height. Write an equation that represents this
relationship. What is the constant of proportionality?
Compound Variation involves both direct and indirect variation. The following is an
example of this.
y varies with x and inversely with z translates to
Example:
Y. Butterworth
y=kx
z
W is varies jointly with T and the square of L and inversely with A.
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