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P. A43: 35-42 S
P. A44: 43-52 S
P. 539: 15-28 S
P. 539: 41-44 S
P. 540: 61-64 S
HW Supplement
Find the least common multiple between 2275
and 2156.
Find the least common multiple between x2 – 4x
and x3 – 8x2 + 16x.
To find the least common multiple of two
algebraic expressions:
1. Factor each expression into primes
2. Multiply the prime factors to the highest
power that they occur in either expression
Objectives:
1. To add, subtract,
multiply, and divide
rational expressions
2. To find the partial
fraction
decomposition of a
rational expression
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Assignment:
P. A43: 35-42 S
P. A44: 43-52 S
P. 539: 15-28 S
P. 539: 41-44 S
P. 540: 61-64 S
HW Supplement
Simplify:
x2
2 x
Multiply:
3x  3x 2 x 2  x  20

2
x  4x  5
3x
Multiplying rational expressions is just like
multiplying any other fractions:
top1
top2
top1  top2


bottom1 bottom2
bottom1  bottom2
Of course, you’ll need to factor and simplify as
needed. One more thing:
a  b  (b  a )
Multiplying rational expressions is just like
multiplying any other fractions:
Divide:
7x
x2  6x
 2
2 x  10 x  11x  30
Dividing rational expressions is just like dividing
any other fractions. Just multiply the first by
the reciprocal of the second:
top1
top2
top1 bottom2



bottom1 bottom2
bottom1 top2
Again, you’ll need to factor and simplify as
needed.
Dividing rational expressions is just like dividing
any other fractions. Just multiply the first by
the reciprocal of the second:
Perform the indicated operation.
2 x 2  10 x x  3
 2
2
x  25 2 x
6 x 2  x  15
2

(3
x
 5 x)
2
4x
Add:
2x
5

x6 x6
Add:
2x
5
x6
Add:
2x
5

x 6 x 3
Adding and subtracting rational expressions is as
painless as adding and subtracting any other
fractions. First you need to get a common
denominator, then you just add the tops.
top1
top2
top1  bottom2  top2  bottom1


bottom1 bottom2
bottom1  bottom2
To get the common denominator, you should
factor first and choose the LCM.
Adding and subtracting rational expressions is as
painless as adding and subtracting any other
fractions. First you need to get a common
denominator, then you just add the tops.
top1
top2
top1  bottom2  top2  bottom1


bottom1 bottom2
bottom1  bottom2
Or you could just multiply each denominator
and simplify twice as much latter.
Perform the indicated operation.
x
5

x 2  x  12 12 x  48
x 1
6

x2  4x  4 x2  4
Simplify:
3
2
3( x  4)
2( x  2)



x2 x4
( x  2)( x  4) ( x  2)( x  4)
3x  12  2 x  4

( x  2)( x  4)

x 8
( x  2)( x  4)

x8
x2  6 x  8
Now suppose that, for
whatever reason, we wanted
to reverse this process. That
would be called Partial
Fraction Decomposition. It’s
like working an addition or a
subtraction problem
backwards to figure out the
original problem.
Write the partial fraction decomposition of:
x 8
x8

x 2  6 x  8 ( x  2)( x  4)
Factor the denominator
x8
A
B


x2  6 x  8 x  2 x  4
Write each factor as a separate
fraction with a generic numerator
x  8  A( x  4)  B( x  2)
Multiply both sides by the LCD.
This is called the Basic Equation.
When all we have are linear factors, solving the basic
equation is a matter of choosing convenient values
for x and solving for A or B.
Solve the basic equation for A:
x  8  A( x  4)  B( x  2)
2  8  A(2  4)  B(2  2)
Let x = −2. Solve for A.
6  2A
3 A
We could have chosen any value for x, since the
equation is true for all values, but when x = −2, B
disappears like mathemagic.
Solve the basic equation for B:
x  8  A( x  4)  B( x  2)
4  8  A(4  4)  B(4  2)
Let x = −4. Solve for A.
4  2B
2  B
Thus, the partial fraction decomposition is:
x8
3
2


x2  6 x  8 x  2 x  4
To decompose a fraction into partial fractions:
1. If the fraction is improper (n ≥ d), divide and
apply the following steps to the remainder.
2. Factor the denominator, not the numerator.
3. For each linear factor in the denominator, write
the partial fractions as:
A
B


a1 x  b1 a2 x  b2

Z
an x  bn
4. Now multiply each side of the equation by the
LCD to obtain the basic equation.
To solve the basic equation:
1. Substitute a convenient value for x to make
one or more of the numerator values (A, B, C,
…) equal zero.
2. Repeat until you find all numerator values.
3. Write the partial fractions using the new
numerator values.
Write the partial fraction decomposition of:
3x 2  x  5
x3  2 x 2  x
When you have repeated linear factors of the
form (ax + b)n:
1. You have to include n partial fractions of the
form: A1
A3
An
A2
ax  b

(ax  b)
2

(ax  b)
3


(ax  b) n
2. When solving the basic equation, substitute
in all known numerator values. Choose other
convenient values of x to find the remaining
numerator values.
Write the partial fraction decomposition of:
x2
x2  4x  3
Write the partial fraction decomposition of:
12
x 3  10 x 2
Write the partial fraction decomposition of:
2 x3  x 2  7 x  7
x2  x  2
Partial fraction decomposition may seem like a
pointless exercise, but it’s not. Your future
calculus teacher wanted me to pre-teach this
since there are some functions that you
cannot integrate unless you decompose them
first into partial fractions.
Also, we only tackled the linear factors; look
forward to the quadratic ones in the future.
Objectives:
1. To add, subtract,
multiply, and divide
rational expressions
2. To find the partial
fraction
decomposition of a
rational expression
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•
•
•
•
Assignment:
P. A43: 35-42 S
P. A44: 43-52 S
P. 539: 15-28 S
P. 539: 41-44 S
P. 540: 61-64 S
HW Supplement