# Section 2.3 Polynomial and Synthetic Division

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```					Section 2.3 Polynomial and
Synthetic Division
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
1. x goes into x3?    x2 times.
x  2x  6
2              2. Multiply (x-1) by x2.
x 1 x  x  4x  6
3  2
4. Bring down 4x.
x x
x  x
3
3   2
2
5. x goes into 2x2? 2x times.
6. Multiply (x-1) by 2x.
0  2 x  4x
2


2x  2x
2
2            8. Bring down -6.
9. x goes into 6x? 6 times.
 0  6x  6     10. Multiply (x-1) by 6.

6 x  6    11. Change sign, Add .

0
Long Division.
x 5
x  3 x  8 x  15
2

Check

 x  3x
2

( x  3)( x  5)               5x  15
 x  5 x  3x  15
2                              
5x  15
 x  8 x  15
2                                  0
x 3x 9
2
Divide.
x  3 x  0 x  0 x  27
3      2

x  27
3                   
 x  3x
3     2

x3                  3x  0 x
2

3x  9 x
2

9 x  27
x  3 x  27
3

9x  27
0
Long Division.
x 2
x  4 x  2x  8
2

Check

x  4x
2

( x  2)( x  4)             2x  8
 x  4x  2x  8
2                        2x  8
2 x  8
 x  2x  8
2                                0
Example
p  2 p  20
2
44
=         p 4
p6                                      p6
p  6 p  2 p  20
2

Check                 p 6p
p  6 p
22

 44 
( p  6)( p  4)  ( p  6)
 p6
    4 p  20
    
 
 4 p  24
 p  4 p  6 p  24  44
2
44
 p  2 p  20
2
p  2 p  20  p  4  44
2

p6                         p6
f ( x)           r ( x)
 q( x) 
d ( x)            d ( x)
p  2 p  20   p  4 p  6 
44
2
( p  6)
p6
p  2 p  20   p  4 p  6  44
2

f ( x)  d ( x) q ( x)  r ( x)
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

f ( x)  d ( x) q ( x)  r ( x)
Where r(x) = 0 or the degree of r(x) is less than
the degree of d(x).
Proper and Improper
f ( x)           r ( x)
 q( x) 
d ( x)           d ( x)
• 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.
Synthetic Division
Divide x4 – 10x2 – 2x + 4 by x + 3
-3        1    0     -10    -2     4
-3      +9     3     -3

1   -3      -1     1      1

x  10 x  2 x  4
4         2
1
 x  3x  x  1 
3    2

x3                             x3
Long Division.

1        -2   -8                     x 1
3
3     3          x  3 x  2x  8
2

 x  3x
2

1   1        -5
x 8

x 3
f ( x)  x  2 x  8
2
5
f (3)  (3)  2(3)  8
2

 9  6 8
 5
The Remainder Theorem
If a polynomial f(x) is divided by x – k, the
remainder is r = f(k).
1x
f ( x)  x  2 x  8
2                   x  3 x  2x  8
2

f (3)  (3)  2(3)  8
2                       x  3x
2

 9  6 8                           x 8
 5                                   
x 3
5
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) = 2x4 + 7x3 – 4x2 – 27x – 18

+2      2     7     -4      -27 -18
4      22      36      18

2    11     18       9       0
Show that (x – 2) and (x + 3) are factors of
f(x) = 2x4 + 7x3 – 4x2 – 27x – 18

+2                    2   7    -4   -27 -18
4    22   36     18

-3                    2   11   18   9
-6 -15    -9
2 x 4  7 x3  4 x 2  27 x  18
2   5    3    0    (2 x 3  11 x 2  18 x  9)( x  2)
(2 x 2  5 x  3)( x  2)( x  3)
Example 6 continued
(2 x  3)(x  1)(x  2)(x  3)
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.
Fun with SYN and the TI-83
f ( x)  x 2  8 x  15    f (3) 

•   Use SYN program to calculate f(-3)
•   [STAT] > Edit
•   Enter 1, 8, 15 into L1, then [2nd][QUIT]
•   Run SYN
•   Enter -3
Fun with SYN and the TI-83
f ( x)  15 x 4  10 x 3  6 x 2  14

• Use SYN program to calculate f(-2/3)
• [STAT] > Edit
• Enter 15, 10, -6, 0, 14 into L1, then
[2nd][QUIT]
• Run SYN
• Enter 2/3
2.3 Homework
• 1-67 odd

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