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Simplifying
Multiplying
Dividing
Grade 9 B – T3 -W 5- 13/14
Simplifying
Remember, denominators
can not = 0.
Now,lets go through the steps to
simplify a rational expression.
Examples of rational expressions
2
4 8 4 7
, ,
3 3 5 9
x y
x x y y
 
  
Simplify:
7x  7
x2
1
Step 1: Factor the numerator and
the denominator completely looking
for common factors.
7x  7  7(x 1)
x2
1  (x 1)(x 1)
Next
7x  7
x2
1

7(x 1)
(x 1)(x 1)
What is the common factor?
x1
Step 2: Divide the numerator and
denominator by the common factor.
7(x 1)
(x  1)(x 1)

7(x 1)
(x  1)(x 1)
1
1
Step 3: Multiply to get your answer.
Answer:
7
x 1
Looking at the answer from the
previous example, what value of x
would make the denominator 0?
x= -1
The expression is undefined when the
values make the denominator equal to 0
How do I find the values
that make an expression
undefined?
Completely factor the original
denominator.
Ex:
2ab(a  2)(b  3)
3ab(a2
 4)
3ab(a2
 4)  3ab(a  2)(a  2)
The expression is undefined when:
a= 0, 2, and -2 and b= 0.
Factor the denominator
Lets go through another example.
3a3
 a4
2a
3
 6a
2
3a3
 a4
2a
3
 6a
2 
a3
(3 a)
2a
2
(a 3)
Factor
out the
GCF
Next
3
2
2 ( 3)
(3 )a
a a
a


  3 factored is 1( 3)a a  
cancel like factors
3
2
1 ( 3)
2 ( 3)
a a
a a
 


1
1
3
2
1( 3)
2 ( 3)
a
a a
a

 

3
2
1
2
a
a
 2
cancel out the like factor a
1
2
a
1
a
answer
What values is the original expression undefined?
Now try to do some on your own.
2
2
3 2
3 2
5 6
1)
9
5 10
2)
6 16
x x
x
x x
x x x
 


 
Also find the values that make
each expression undefined?
Multiplying and
Dividing
Multiplying and Dividing
Rational Expressions
Remember that a rational number
can be expressed as a quotient of
two integers. A rational expression
can be expressed as a quotient of
two polynomials.
Remember how to multiply fractions:
First you multiply the numerators
then multiply the denominators.
5 2
:
6 20
Ex 
10 1
120 12

5 2
6 20

The same method can be used
to multiply rational expressions.
Ex:
4a2
5ab
3 
3bc
12a
3 
4  a  a 3  b c
5 a  b  b b 12  a  a a
11 1 1 1
1 1 1 1

c
5b2
a2
Let’s do another one.
Ex:
x3
 3x2
x
2
 5x  6

x2
10x  9
x
2
 6x  27
Step #1: Factor the numerator
and the denominator.
x2
(x  3)
(x  6)(x 1)

(x 1)(x  9)
(x  9)(x 3)
Next
Step #2: Divide the numerator and
denominator by the common factors.
x2
(x  3)
(x  6)(x 1)

(x 1)(x  9)
(x  9)(x 3)
1
1
1
1
1
1
Step #3: Multiply the numerator
and the denominator.
x2
x  6
Remember how to divide fractions?
Multiply by the reciprocal of the
divisor. 4
5

16
25

4
5

25
16

4  25
516
1
1
5
4

5
4
Dividing rational expressions uses
the same procedure.
Ex: Simplify
y  2
y
2
10y  24

y2
 2y
y
2
 2y  8
y  2
y
2
10 y  24

y2
 2y
y
2
 2y  8

y  2
y
2
10y  24

y2
 2y  8
y
2
 2y

y  2
(y 12)(y  2)

(y  4)(y  2)
y(y  2)

1 1
1 1
Next
4
( 12)
y
y y


Now you try to simplify the
expression:
x  3
x
2
 4x 12

2x2
 6x
x  2
Choose the correct answer .
Answer:
1
2x(x  6)
Now try these on your own.
1)
x + 3
2x3
 2x2

x2
 7x  6
x2
10x  21
2)
3x  6
7x  7

5x 10
14x 14

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Rational Expressions Multiplying and Dividing Grade 9

  • 3. Remember, denominators can not = 0. Now,lets go through the steps to simplify a rational expression. Examples of rational expressions 2 4 8 4 7 , , 3 3 5 9 x y x x y y     
  • 4. Simplify: 7x  7 x2 1 Step 1: Factor the numerator and the denominator completely looking for common factors. 7x  7  7(x 1) x2 1  (x 1)(x 1) Next
  • 5. 7x  7 x2 1  7(x 1) (x 1)(x 1) What is the common factor? x1 Step 2: Divide the numerator and denominator by the common factor.
  • 6. 7(x 1) (x  1)(x 1)  7(x 1) (x  1)(x 1) 1 1 Step 3: Multiply to get your answer. Answer: 7 x 1
  • 7. Looking at the answer from the previous example, what value of x would make the denominator 0? x= -1 The expression is undefined when the values make the denominator equal to 0
  • 8. How do I find the values that make an expression undefined? Completely factor the original denominator.
  • 9. Ex: 2ab(a  2)(b  3) 3ab(a2  4) 3ab(a2  4)  3ab(a  2)(a  2) The expression is undefined when: a= 0, 2, and -2 and b= 0. Factor the denominator
  • 10. Lets go through another example. 3a3  a4 2a 3  6a 2 3a3  a4 2a 3  6a 2  a3 (3 a) 2a 2 (a 3) Factor out the GCF Next
  • 11. 3 2 2 ( 3) (3 )a a a a     3 factored is 1( 3)a a   cancel like factors 3 2 1 ( 3) 2 ( 3) a a a a     1 1 3 2 1( 3) 2 ( 3) a a a a    
  • 12. 3 2 1 2 a a  2 cancel out the like factor a 1 2 a 1 a answer What values is the original expression undefined?
  • 13. Now try to do some on your own. 2 2 3 2 3 2 5 6 1) 9 5 10 2) 6 16 x x x x x x x x       Also find the values that make each expression undefined?
  • 15. Multiplying and Dividing Rational Expressions Remember that a rational number can be expressed as a quotient of two integers. A rational expression can be expressed as a quotient of two polynomials.
  • 16. Remember how to multiply fractions: First you multiply the numerators then multiply the denominators. 5 2 : 6 20 Ex  10 1 120 12  5 2 6 20 
  • 17. The same method can be used to multiply rational expressions. Ex: 4a2 5ab 3  3bc 12a 3  4  a  a 3  b c 5 a  b  b b 12  a  a a 11 1 1 1 1 1 1 1  c 5b2 a2
  • 18. Let’s do another one. Ex: x3  3x2 x 2  5x  6  x2 10x  9 x 2  6x  27 Step #1: Factor the numerator and the denominator. x2 (x  3) (x  6)(x 1)  (x 1)(x  9) (x  9)(x 3) Next
  • 19. Step #2: Divide the numerator and denominator by the common factors. x2 (x  3) (x  6)(x 1)  (x 1)(x  9) (x  9)(x 3) 1 1 1 1 1 1
  • 20. Step #3: Multiply the numerator and the denominator. x2 x  6 Remember how to divide fractions?
  • 21. Multiply by the reciprocal of the divisor. 4 5  16 25  4 5  25 16  4  25 516 1 1 5 4  5 4
  • 22. Dividing rational expressions uses the same procedure. Ex: Simplify y  2 y 2 10y  24  y2  2y y 2  2y  8
  • 23. y  2 y 2 10 y  24  y2  2y y 2  2y  8  y  2 y 2 10y  24  y2  2y  8 y 2  2y  y  2 (y 12)(y  2)  (y  4)(y  2) y(y  2)  1 1 1 1 Next 4 ( 12) y y y  
  • 24. Now you try to simplify the expression: x  3 x 2  4x 12  2x2  6x x  2
  • 25. Choose the correct answer .
  • 26. Answer: 1 2x(x  6) Now try these on your own. 1) x + 3 2x3  2x2  x2  7x  6 x2 10x  21 2) 3x  6 7x  7  5x 10 14x 14