Dividing
Polynomials
9-12-’12
Factoring Polynomials, Roots
of Real Numbers
9-13-’12
CCSS:N.RN.2 & A.APR.1
N.RN.2 REWRITE expressions involving
radicals and rational exponents using the
properties of exponents
A.APR.1 UNDERSTAND that polynomials
FORM a system analogous to the integers,
namely, they ARE CLOSED under the operations
of addition, subtraction, and multiplication; ADD,
SUBTRACT, and MULTIPLY polynomials.

Essential Question(s):
How do I divide polynomial expressions?
How do I factor a polynomial expression?
How do I interpret the parts of a factored
expression in context of the variables?
How do I evaluate roots of real numbers?
How do I rewrite roots of real numbers as
rational exponents?
Standards for Mathematical Practice








1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments and critique the reasoning of others.
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning.
Simple Division dividing a polynomial by a monomial
6r s  3rs  9r s
1.
3rs
2 2
2
2
6r s
3rs
9r s



3rs
3rs
3rs
2 2
2
2
 2rs  s  3r
Simplify
3a b  6a b  18ab
2.
3ab
2
3 2
2
3 2
3a b 6a b 18ab



3ab
3ab
3ab
2
 a  2a b  6
Simplify
12 x y  3 x
3.
3x
2
2
12x y 3x


3x
3x
 4xy 1
Long Division divide a polynomial by a polynomial
•Think back to long division from 3rd grade.
•How many times does the divisor go into the
dividend? Put that number on top.
•Multiply that number by the divisor and put the
result under the dividend.
•Subtract and bring down the next number in the
dividend. Repeat until you have used all the
numbers in the dividend.
x  5 x  24
4.
x3
2

x- 8
2
x  3 x  5x  24
-(x2 + 3x)
2
x /x = x
- 8x - 24
-(- 8x - 24)
-8x/x = -8
0
5.
h
3
 11h  28  h  4 
48
2
h + 4h + 5 
h4
1

3
2
h  4 h  0h  11h  28
3
h /h
=
2
h
2
4h /h
-(h3
-
2
4h )
2
4h 2 - 11h
-(4h - 16h)
5h
+
28
= 4h
-(5h - 20 )
5h/h = 5
48
Synthetic Division divide a polynomial by a polynomial
To use synthetic division:
•There must be a coefficient for
every possible power of the variable.
•The divisor must have a leading
coefficient of 1.
Ex6 :
5x
4

 4 x  x  6  ( x  3)
2
5x
4

 4 x  x  6  ( x  3)
2
Step #1: Write the terms of the
polynomial so the degrees are in
descending order.
4
3
2
5x  0x  4x  x  6
Since the numerator does not
contain all the powers of x,
3
you must include a 0 for the x .
5x
4

 4 x  x  6  ( x  3)
2
Step #2: Write the constant r of the
divisor x-r to the left and write
down the coefficients.
4
3
2
5x  0x  4x  x  6
3
5
0
-4
1
Since the divisor is x-3, r=3
6
5x
4

 4 x  x  6  ( x  3)
2
Step #3: Bring down the first
coefficient, 5.
3
5
5
0
-4
1
6
5x
4
2
 4x  x  6
  ( x  3)
Step #4: Multiply the first
coefficient by r, so 3  5  15
and place under the second
coefficient then add.
3
5
0
15
5
15
-4
1
6
5x
4
2
 4x  x  6
  ( x  3)
Step #5: Repeat process multiplying
the sum, 15, by r; 15  3  45
and place this number under the
next coefficient, then add.
3
5
5
0
-4
15
45
15
41
1
6
 5x
4
2
 4x  x  6
  ( x  3)
Step #5 cont.: Repeat the same procedure.
Where did 123 and 372 come from?
3
5
5
0
-4
1
6
15
45
123 372
15
41 124 378
 5x
4
2
 4x  x  6
  ( x  3)
Step #6: Write the quotient.
The numbers along the bottom are coefficients
of the power of x in descending order, starting
with the power that is one less than that of the
dividend.
3
5
5
0
-4
1
6
15
45
123 372
15
41 124 378
 5x
4
2
 4x  x  6
  ( x  3)
The quotient is:
378
5x  15x  41x  124 
x3
3
2
Remember to place the
remainder over the divisor.
Ex 7:
5x
5
 21x  3x  4x  2x  2  x  4
4
3
2
Step#1: Powers are all accounted
for and in descending order.
Step#2: Identify r in the divisor.
Since the divisor is x+4, r=-4 .
4
5
 21 3
4
2 2
 5 x  21x  3 x  4 x  2 x  2    x  4 
Step#3: Bring down the 1st coefficient.
Step#4: Multiply and add.
Step#5: Repeat.
5
4
4
5
3
2
 21 3
20
-1
4
1
4
-4
0
2 2
0 8
-2 10
-5
10
4
3
2
5 x  x  x  2 
x4
Ex 8:
6x
2
 2x  4 2x  3
Notice the leading coefficient of the
divisor is 2 not 1.
We must divide everything by 2 to
change the coefficient to a 1.
2
6x
2x 4  2x 3 

     

 2
2
2 
2 2
3 
2

 3x  x  2 x 
 2 
6x
3
2
2

 2 x  4   2 x  3
3
1
 
2

2
2
3
9
2
7
2
21
4
29
4

8
4


6 x  2 x  4   2 x  3
2
3x 
7
2
 3x 
29

7
2
4
x

3
 3x 
2
29
3

4 x  
2

7
2

29
4

1
x
3
*Remember we
2
cannot have
complex fractions we must simplify.
 3x 
7
2

29
4x  6
Ex 9:
x
3
 x  2x  7  2x  1
2
 x x 2x 7   2x 1 



 
 
2
2 2  2 2
 2
3
1
2
2

1
2
1 7
2
Coefficients
3
2
x

x
 2 x  7    2 x  1

7 
1
1 3 1 2
 x  x  x   x  
2
2 
2
2
1
2
1

2


2
1
4
2

1

8
8



2
7
16
7
4
1
8
7
16
4
8

56

1
1
1

2
49
16
A “Difference of Squares” is
a binomial (*2 terms only*)
and it factors like this:
2
2
a  b  (a  b)(a  b)
Factoring a polynomial
means expressing it as a
product of other
polynomials.
Factoring Method #1
Factoring polynomials with a
common monomial factor (using
GCF).
**Always look for a GCF before
using any other factoring method.
Steps:
1. Find the greatest common factor
(GCF).
2. Divide the polynomial by the GCF.
The quotient is the other factor.
3. Express the polynomial as the product
of the quotient and the GCF.
Example :
6c d  12c d  3cd
3
2
2
GCF  3cd
Step 1:
Step 2: Divide by GCF
3
2
2
(6c d  12c d  3cd)  3cd 
2
2c  4cd  1
The answer should look like this:
3
2
2
Ex: 6c d 12c d  3cd
2
 3cd(2c  4cd  1)
Factor these on your own
looking for a GCF.
1. 6x  3x  12x  3 x  2 x
3
2
2
2. 5x  10x  35
2
 x  4
 5 x2  2 x  7
3. 16x y z  8x y z  12xy z
3 4
2
2 3
 4 xy z  4 x y  2 xz  3 yz 
2
2
2
2
3 2
Factoring Method #2
Factoring polynomials that are a
difference of squares.
To factor, express each term as a
square of a monomial then apply
2
2
the rule... a  b  (a  b)(a  b)
2
Ex: x 16 
2
2
x 4 
(x  4)(x  4)
Here is another example:
1 2
x  81 
49
2
1 x  92  1 x  91 x  9
7
7

7 
Try these on your own:
1. x  121
2
  x  11 x  11
2. 9y  169x
2
2
  3 y  13 x  3 y  13 x 
3. x  16   x  2  x  2   x  4 
4
Be careful!
2
Sum and Difference of Cubes:
a  b  a  ba  ab  b
3
3
a b
3
3

 a  ba  ab  b 
2
2
2
2
Write each monomial as a cube and apply
either of the rules.
Example :
Rewrite as cubes
3
3
x  64  (x  4 )
3
Apply
the
rule
for
sum
3
3
2
2
a cubes:
 b  a  b a  ab  b
of

2
2
 (x  4)(x  x 4  4 )
2
 (x  4)(x  4x  16)

Rewrite as cubes
3
3
3
Ex: 8y  125  ((2y)  5 )
Apply the rule for difference
3
3
2
2
aof cubes:
 b  a  b a  ab  b



 2y  5 2y  2y  5  5
2
2
 2y  54y  10y  25
2

Factoring Method #3
Factoring a trinomial in the form:
2
ax  bx  c
Factoring a trinomial:
2
ax  bx  c
1. Write two sets of parenthesis, ( )( ).
These will be the factors of the
trinomial.
2. Product of first terms of both binomials
must equal first term of the trinomial.
2
( ax )
Nex
Factoring a trinomial:
2
ax  bx  c
3. The product of last terms of both
binomials must equal last term of the
trinomial (c).
4. Think of the FOIL method of
multiplying binomials, the sum of the
outer and the inner products must
equal the middle term (bx).
Example :
x  6x  8
2
x
 x 
x -2 x -4 
Factors of +8:
1&8
2&
4
x x  x
2
O + I = bx
?1x + 8x =
9x
2x + 4x =
6x - 8x =
-1x
-9x
-2x - 4x =
-6x
x  6x  8  (x  2)(x  4)
2
Check your answer by
using FOIL
F2
O
I
L
(x  2)(x  4)  x  4x  2x  8
2
 x  6x  8
Lets do another example:
2
6x 12x 18
Don’t Forget Method #1.
Always check for GCF before you do
2
anything
6(x  2x
 3) else.Find a
GCF
6(x  3)(x 1) Factor
trinomial
When a>1 and c<1, there may be
more combinations to try!
Example :
6 x  13x  5
2
Step 1:
2
Find the factors of 6x :
3x  2x
6x  x
Example : 6 x  13 x  5
2
Step 2:
Find the factors of -5:
5 -1
-5 1
 -1 5 


1 -5 
Order can make
a difference!
Example : 6 x  13 x  5
2
Step 3: Place the factors inside the
parenthesis until O + I = bx.
Try:
F2
6x  1x  5
O
I
L
6x  30x  x  5
O + I = 30 x - x = 29x
This
doesn’t
work!!
Example : 6 x  13 x  5
2
Switch the order of the second terms
and try again.
6x  5x  1
F2
O
I
L
6x  6x  5x  5
O + I = -6x + 5x = -x
This
doesn’t
work!!
Try another combination:
Switch to 3x and 2x
(3x 1)(2x  5)
F2
O
I
L
6x  15x  2x  5
O+I = 15x - 2x = 13x IT WORKS!!
6x  13x  5  (3x 1)(2x  5)
2
Factoring Technique #3
continued
Factoring a perfect square trinomial
in the form:
a  2ab  b  (a  b)
2
2
a  2ab  b  (a  b)
2
2
2
2
Perfect Square Trinomials can be
factored just like other trinomials (guess
and check), but if you recognize the
perfect squares pattern, follow the
formula!
a  2ab  b  (a  b)
2
2
a  2ab  b  (a  b)
2
2
2
2
2
Ex: x  8x 16
x 
4 
2
2
a
b
Does the
middle term 2  x  4  8x
fit the
Yes,
the
factors
are
(a
+
pattern, 2
2
2 b) :
2ab? x  8x  16   x  4
2
Ex: 4x 12x  9
2x 
3
2
2
a
b
Does the
3
2x

 12x
2

middle term
fit the
Yes,
the
factors
are
(a
pattern, 2
2
2 b) :
2ab?
4x  12x  9  2x  3
Factoring Technique #4
Factoring By Grouping
for polynomials
with 4 or more terms
Factoring By Grouping
1. Group the first set of terms and
last set of terms with parentheses.
2. Factor out the GCF from each group
so that both sets of parentheses
contain the same factors.
3. Factor out the GCF again (the GCF
is the factor from step 2).
3
Example b  3b
1:
Step 1: Group
 b  3b
3
2
2
 4 b  12
 4b  12 
Step 2: Factor out GCF from each group
 b b  3  4b  3
2
Step 3: Factor out GCF again
 b  3b  4
2
Example 2 x  16 x  8 x  64
3
2
2:
 2 x  8x  4x  32
3
2


3
2
 2x  8x  4x  32
 2x x  8  4x  8
2
 2x  8x  4 
2
 2x 8x  2x  2
Try these on your own:
2
1. x  5x  6
2
2. 3x  11x  20
3
3. x  216
4. 8x  8
3
5. 3x  6x  24x
3
2
Answers:
1. (x  6)(x  1)
2. (3x  4)(x  5)
2
3. (x  6)(x  6x  36)
4. 8(x  1)(x  x  1)
2
5. 3x(x  4)(x  2)
Roots of Real
Numbers and Radical
Expressions
Application
 A carton shaped like a cube has
sides 14 inches long. The
volume of the carton is 14  14
 14 or 143 cubic inches (in3).
Can you see why we call 14 to
the third power 14 cubed? Why
do you think we call raising a
number to the second power
squaring the number? Why do
you suppose that there are not
special names for raising a
number to any other power?
SLIDE 65
Definition of
th
n
Root
For any real numbers a and b
and any positive integers n,
if an = b,
then a is the nth root of b.
** For a square root the value of n is 2.
Notation
radical
index
4
81
radicand
Note: An index of 2 is understood but not
written in a square root sign.
4
Simplify
81
To simplify means to find x
in the equation:
4
x = 81
Solution:
4
81=
3
Principal Root
The nonnegative root of a number
64
Principal square root
 64
Opposite of principal
square root
 64
Both square roots
Summary of Roots
b
The n th root of b
n b >0 b <0 b =0
n
even
odd
one + root
one - root
one + root
no - roots
no real
roots
no + roots
one - root
one real
root, 0
Examples


1.  169 x
4
2. -
13 x 
2 2


8
x
3


4
 13x
  8 x  3 
2 2
   8 x  3
2
2
Examples
3.
4.
3

125 x
3
6
3
m n 
3
3
5 x 
3
2
3
 5x
2
 mn  mn
3
Taking
th
n
roots of variable
expressions
Using absolute value signs
If the index (n) of the radical is
even, the power under the radical
sign is even, and the resulting
power is odd, then we must use an
absolute value sign.
Examples
Even
Odd
Even
1.
4
 an   an
4
Even
2.
6
 xy 
6
2
Even
xy
Odd
2
Even
Odd
Even
2
3.
x
6
 x
3
Odd
Even
4.
6
3  y 
2 18
 3 - y
3
2
Even
 3 - y

3
2
Product Property of
Radicals
For any numbers a and
b where a 0and b 0,
ab  a  b
Product Property of
Radicals Examples
72 
36 2  36 2
6 2
48  16 3  16 3
4 3
Examples:
1.
30a  a  30
34
34
 a
17
2.
30
54x y z  9x y z  6yz
4 4 6
4 5 7
 3x y z
2
2
3
6 yz
Examples:
3.
3
54a b  27a b  2b
3 7
3
3 7
3
 3ab  2b
2
4.
3
60xy  4 y  15xy
3
2
 2 y 15xy
Quotient Property of
Radicals
For any numbers a and
b where a 0and b 0,
a

b
a
b
Examples:
1.
7

16
32

2.
25
7
7

4
16
32
25

32 4 2

5
5
Examples:
3.
4.
48
3

45

4
48

3
16  4
45
45 3 5

2
2
4

Rationalizing the
denominator
Rationalizing the denominator means
to remove any radicals from the
denominator.
Ex: Simplify
5

3
5
3

3
3

5 3
9

5 3

3
15
3
Simplest Radical Form
•No perfect nth power factors
other than 1.
•No fractions in the radicand.
•No radicals in the denominator.
Examples:
1.
5

4
20 8
5
5

2
4
8
2.
 10  10 4  10 2
2
2 2
 20
Examples:
5
2
5 2
5
2

3.



4
2 2 2 2 4
2 2
5 2
5 4 5 7x 4 35x
4. 4



2
7x
7x
7x
49x
4 35x

7x
Adding radicals
We can only combine terms with radicals
if we have like radicals
6 7 5 7  3 7
 6 5 3 7  8 7
Reverse of the Distributive Property
Examples:
1. 2 3 + 5 + 7 3 - 2
= 2 3 + 7 3 + 5- 2
= 9 3+3
Examples:
2. 5 6  3 24  150
= 5 6  3 4 6  25 6
= 5 6 6 6  5 6
=4 6
Multiplying radicals Distributive Property
 2  4 3
3

3 2  3 4 3

6  12
Multiplying radicals - FOIL

3 5
F


24 3
O
3 2  3 4 3
I

L
 5 2  5 4 3

6  12 10  4 15
Examples:



1. 2 3  4 5 3  6 5
O
F
 2 3 3  2 3 6 5
L
I
4 5  3  4 5  6 5
 6  12 15  4 15  120
 16 15  126
Examples:

5 4  2 7
= 5 2 2 75 2 2 7
O
F
2. 5 4  2 7
 1010 10 2 7
I
L
2 7 10 2 7  2 7
 100 20 7  20 7  4 49
 100 4 7  72
Conjugates
Binomials of the form
a b  c d anda b  c d
where a, b, c, d are rational
numbers.
Ex:
5  6  Conjugate:
56
3  2 2  Conjugate: 3  2 2
What is conjugate of
2 7  3?
Answer: 2 7  3
The product of conjugates is a
rational number. Therefore, we can
rationalize denominator of a fraction
by multiplying by its conjugate.
Examples:
32 35

1.
35 35
3  3  5 3  2 3  2 5

2
2
3 5
3  7 3  10 13 7 3


22
3 25
 
Examples:
1 2 5 6  5

2.
6 5 6  5

6  5  12 5  10
6 
2
 
2
5
16 13 5

31