Dividing Polynomials

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Transcript Dividing Polynomials

Dividing
Polynomials
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