This preview shows pages 1–3. Sign up to view the full content.
Preparation for Chapter 4
1 Synthetic Division
If a polynomial p(x) is divided by (x – c) we can apply synthetic division:
Let’s divide x
3
– 4x
2
– 5 by x – 3
Step 1: Write the dividend in descending powers of x. Copy the coefficients; write a zero
for any missing power.
1 4 0 5
Step 2: Create the following structure:
3
1
!
4 0
!
5 Row 1
Row 2
Row 3
First row, we place: 1 4 0 5
Second row, to the left of the symbol, we place 3 since the divisor is x – 3.
Step 3: Bring the 1 down two rows, and enter it in row 3.
3
1
!
4 0
!
5
"
1
Step 4: Multiply the 1, in row 3, by 3, and place the result 3, in row 2 underneath the 4.
3
1
!
4 0
!
5
"
3
1
#
3
!
Step 5: Add the entries in row 1 and row 2 (4 + 3 = 1) and enter the sum in row 3.
3
1
!
4 0
!
5
"
3
1
#
3
!
!
1
Step 6: Multiply the 1, in row 3, by 3, and place the result 3, in row 2 underneath the 0.
3
1
!
4
0
!
5
"
3
!
3
1
#
3
!
!
1
#
3
!
Step 7: Add the entries in row 1 and row 2 (0 + 3 = 3) and enter the sum in row 3.
3
1
!
4
0
!
5
"
3
!
3
1
#
3
!
!
1
#
3
!
!
3
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3
1
!
4
0
!
5
"
3
!
3
!
9
1
#
3
!
!
1
#
3
!
!
3
#
3
!
Step 9: add the entries in row 1 and row 2 (5 + 9 = 14) and enter the sum in row 3.
3
1
!
4
0
!
5
"
3
!
3
!
9
1
#
3
!
!
1
#
3
!
!
3
#
3
!
!
14
End of the process.
The numbers 1 1 and 3 are the coefficients of a second degree polynomial that is the
quotient (x
2
– x – 3), the number 14 is the remainder, so
x
3
 4x
2
– 5 = (x  3)(x
2
– x – 3) + (14)
Example
:
Use synthetic division to divide (2x
5
+ 5x
4
– 2x
3
+ 2x
2
– 2x + 3) by (x + 3)
The divisor is x + 3 = x – (3), so we place 3 to the left of the symbol.
Remember, we add each entry in row 1 to the corresponding entry in row 2 and the sum
is place in row 3.
!
3
2
5
!
2
2
!
2
3 Row 1
"
!
6
3
!
3
3
!
3 Row 2
2
# !
3
(
)
!
!
1
# !
3
(
)
!
1
# !
3
(
)
!
!
1
# !
3
(
)
!
1
# !
3
(
)
!
0 Row 3
The remainder is 0, the quotient is the polynomial 2x
4
– x
3
+ x
2
– x + 1, so
2x
5
+ 5x
4
– 2x
3
+ 2x
2
– 2x + 3) = (x + 3)( 2x
4
– x
3
+ x
2
– x + 1)
2 Roots or Zeros of a Polynomial
Given a polynomial P(x), we say that a number c is a root or zero of the polynomial if
P(c) = 0.
The Factor Theorem
:
The number c is a root or zero of the polynomial P(x) if and only if (x – c) is a factor of
P(x). (P(x) = (x – c) Q(x) where degree Q(x) equals degree P(x) minus 1)
Example
:
Show that x – 1 is a factor of P(x) = 2x
3
– x
2
+ 2x – 3.
Applying the theorem, we have to show that P(1) = 0.
P(1) = 2(1)
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 STAFF

Click to edit the document details