polynomial_and_synthetic_dfdgfgivision.ppt

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1. Section 2.3 Polynomial and
Synthetic Division

2. What you should learn
•How to use long division to divide
polynomials by other polynomials
•How to use synthetic division to divide
polynomials by binomials of the form
(x – k)
•How to use the Remainder Theorem and the
Factor Theorem

3. 641
23
 xxxx
2
x
1. x goes into x
3
? x
2
times.
2. Multiply (x-1) by x
2
.
23
xx
2
20x x4
4. Bring down 4x.
5. x goes into 2x
2
? 2x

times.
x2
6. Multiply (x-1) by 2x.
xx22
2

x60
8. Bring down -6.
6
9. x goes into 6x?
6
66x
0
3. Change sign, Add.
7. Change sign, Add
6

times.
11. Change sign, Add .
10. Multiply (x-1) by 6.
3 2
x x 
2
2 2x x 
6 6x 

4. Long Division.
1583
2
 xxx
x
xx3
2

155x
5
155x
0
)5)(3( xx
Check
1535
2
 xxx
158
2
 xx
2
3x x 
5 15x 

5. Divide.
3
27
3
x
x


3
3 27x x 
3 2
3 0 0 27x x x x   
2
x
3 2
3x x
3 2
3x x 
2
3 0x x
3x
2
3 9x x
2
3 9x x 
9 27x
9
9 27x9 27x 
0

6. Long Division.
824
2
 xxx
x
xx4
2

82x
2
82x
0
)4)(2( xx
Check
824
2
 xxx
82
2
 xx
2
4x x 
2 8x 

7. Example
2026
2
 ppp
p
pp6
2

204p
4
244p
44










6
44
)6()4)(6(
p
ppp
Check
442464
2
 ppp
202
2
 pp
6
202
2


p
pp
6
44


p
2
6p p 
4 24p 
=

8. 202
2
pp
6
202
2


p
pp
6
44
4


p
p
 )6(
6
44
64 

 p
p
pp
4464  pp202
2
pp
)(
)(
)(
)(
)(
xd
xr
xq
xd
xf

)()()()( xrxqxdxf 

9. The Division Algorithm
If f(x) and d(x) are polynomials such that d(x) ≠ 0,
and the degree of d(x) is less than or equal to the
degree of f(x), there exists a unique polynomials
q(x) and r(x) such that
Where r(x) = 0 or the degree of r(x) is less than
the degree of d(x).
)()()()( xrxqxdxf 

10. Proper and Improper
•Since the degree of f(x) is more than or equal
to d(x), the rational expression f(x)/d(x) is
improper.
•Since the degree of r(x) is less than than d(x),
the rational expression r(x)/d(x) is proper.
)(
)(
)(
)(
)(
xd
xr
xq
xd
xf


11. Synthetic Division
Divide x
4
– 10x
2
– 2x + 4 by x + 3
10-10-24-3
1
-3
-3
+9
-1
3
1
-3
1



3
4210
24
x
xxx
3
1


x
13
23
 xxx

12. Long Division.
823
2
 xxx
x
xx3
2

8x
1
3x
582)(
2
 xxxf
xx3
2

3x
)3(f 8)3(2)3(
2

869
5
1-2-83
1
3
1
3
-5

13. The Remainder Theorem
If a polynomial f(x) is divided by x – k, the
remainder is r = f(k).
82)(
2
 xxxf
)3(f 8)3(2)3(
2

869
5
823
2
 xxx
x
xx3
2

8x
1
3x
5
xx3
2

3x

14. The Factor Theorem
A polynomial f(x) has a factor (x – k) if and only
if f(k) = 0.
Show that (x – 2) and (x + 3) are factors of
f(x) = 2x
4
+ 7x
3
– 4x
2
– 27x – 18
27-4-27-18+2
2
4
11
22
18
36
9
18
0

15. Example 6 continued
Show that (x – 2) and (x + 3) are factors of
f(x) = 2x
4
+ 7x
3
– 4x
2
– 27x – 18
27-4-27-18+2
2
4
11
22
18
36
9
18
-3
2
-6
5
-15
3
-9
0
1827472
234
 xxxx
)2)(918112(
23
 xxxx
)3)(2)(352(
2
 xxxx
)3)(2)(1)(32(  xxxx

16. Uses of the Remainder in Synthetic
Division
The remainder r, obtained in synthetic division
of f(x) by (x – k), provides the following
information.
1.r = f(k)
2.If r = 0 then (x – k) is a factor of f(x).
3.If r = 0 then (k, 0) is an x intercept of the
graph of f.

17. Fun with SYN and the TI-83
•Use SYN program to calculate f(-3)
•[STAT] > Edit
•Enter 1, 8, 15 into L1, then [2
nd
][QUIT]
•Run SYN
•Enter -3
158)(
2
 xxxf )3(f

18. Fun with SYN and the TI-83
•Use SYN program to calculate f(-2/3)
•[STAT] > Edit
•Enter 15, 10, -6, 0, 14 into L1, then [2
nd
]
[QUIT]
•Run SYN
•Enter 2/3
1461015)(
234
 xxxxf

19. 2.3 Homework
•1-67 odd