Simplifying rational expressions

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Multiplying and dividing Rational
functions with factoring first.
There are 2
to doing these problems:
Factoring
And
Simplifying numerators with
denominators
Lets start with simplifying a rational
expression.
(x+6)
Factor the
numerator:
C ancel out the
common terms in
the numerator
and denominator.
Factor the
denominator:
(x-4)
x  2 x  24
2
x  7x  6
2
(x+6)
Original
problem
(x+1)
The answer
is…
x4
x 1
To simplify a
binomial they
must be the exact
same. Monomial
variables subtract
exponents.
Numbers divide
or reduce.
Common
term
Difference
of Squares
Factor
Grouping
Trinomial
Multiplying a rational expression.
Factor all the
numerators :
2x(x-5)
(x+3)
2 x  10 x x  3

2
2
x  25
2x
2
(x+5)(x-5)
Factor all the
denominators:
2
2x
Multiplying a rational expression.
2x(x-5)
(x+3)
2 x  10 x x  3

2
2
x  25
2x
2
(x+5)(x-5)
Cancel out or simplify
the common terms in
the numerator and
denominator no matter
where they are.
2
2x
x
The answer
is…
x3
x( x  5)
*** very important an x by
itself can not cancel with
an x that is being added
or subtracted to a
number
2
4x
2
20 x  12 x
2
4x
4 x (5 x  3)
x
5x  3
2 x  10 x
2
3x  16 x  5
2
2 x ( x  5)
(3 x  1)( x  5)
2x
3x  1
x 2  3x  4 2 x 2  4 x
 2
2
x  4x  4 x  4x  3
( x  4)( x  1)
2 x( x  2)

( x  2)( x  2) ( x  3)( x  1)
2 x( x  4)
( x  2)( x  3)
Practice
x - 9x + 8 4x
· 2
2
3x - 24x x -1
2
3
x 2  3x  10 x 2  10 x  21

2
x  2 x  15
Dividing Rational expressions
You will flip over this one.
Literally you flip the rational that comes AFTER the division sign
then…
The division becomes
multiplication and you do the
same thing as the last problems.
4x
x  2x
 2
5 x  20 x  6 x  8
2
4x
x  6x  8
 2
5 x  20
x  2x
2
4x
( x  4)( x  2)

5( x  4)
x( x  2)
4
5
3x  13x  4 4 x  16

2
x 4
x2
2
(3x  1)( x  4) ( x  2)

( x  2)( x  2) 4( x  4)
3x  1
4( x  2)
Practice
x -13x +36 x + 3x - 4x -12
¸ 3
4
2
2
x -14x - 32 x - 4x + 2x - 8
4
2
3
2
Adding and Subtracting Rational
Functions
You still get to… … but only the denominator
Factor
Basically you find the
Least common
denominator (which will
include factors) and
multiply the numerators
by the missing factors.
2
+
5
2
5
 2
2
3x 3x
2
3x
LCD since both denominators are
the same
7
2
3x
This rational
needs to multiply
by x.
This rational
needs to multiply
by (x-4)
1
x
1
x
 2
2
3x 3x  12 x
3x2
3x(x-4)
Multiply each
numerator by
what the
denominator is
missing.
LCD: 3x2 (x-4) must contain all of the
denominators but nothing extra
And the
solution is…
1( x  4)  x( x) x  4  x
 2

2
3x ( x  4)
3x ( x  4)
2
x  x4
2
3x ( x  4)
2
How do I get the LCD?
First I find the smallest number that all the coeficiants
will divide into evenly. Ex. 4 and 10 would be 20
because that is the smallest number that 4 and 10
divide into evenly.
Next I take any base variable that is any denominator
and use the one with the largest exponent Ex. x, x2 ,
and x3 the LCD is x3
Lastly any binomial in the denominator I do the same
as a variable. Ex (x + 2)(x + 2) and (x + 2)(x – 1)
the LCD is (x + 2)2 (x – 1) because there are 2 (x + 2)’s
in one denominator.
3 1

4x 7
lcd : 28 x
3(7) 1(4 x)

28 x 28 x
21  4 x
28 x
x 1
6
 2
2
x  4x  4 x  4
factor :
( x  2)( x  2) ( x  2)( x  2)
lcd : ( x  2)( x  2)( x  2)
( x  1)( x  2)
6( x  2)

2
( x  2) ( x  2) ( x  2) 2 ( x  2)
x 2  x  2  6 x  12
( x  2) 2 ( x  2)
x 2  7 x  14
( x  2) 2 ( x  2)
Practice
1 1
+
x y
x -3
3
+ 2
2
x - 2x - 3 x + 3x + 2

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