Theory of Polynomials
Summary 1
Synthetic Division
Synthetic Division can be used when dividing a polynomial P(x) by a binomial of the
form x –c
Example 1: Divide the polynomial x5  3x 4  15x3  4x 2  6x  8 by x  2
Step 1 - Write the number c and the coefficients of the polynomial on the same line.
2 |
1
-3
15
-4
-6
8
Step 2 - Bring down the first coefficient two spaces and multiply by c. Place the
product under the second coefficient.
2 |
1
-3
2
15
-4
-6
8
1
Step 3 - Combined the second coefficient with the product
2 |
1
-3
15
-4
-6
8
2
1
-1
Step 4 - Multiply the result obtained in step 3 by c and place the product under the third
coefficient. Combine the product with the third coefficient
2 |
1
-3
15
-4
-6
8
2
-2
1
-1
13
Step 5 - Multiply the result obtained in step 4 by c and place the product under the
fourth coefficient. Combine the product with the fourth coefficient
2 |
1
-3
15
-4
-6
8
2
-2
26
1
-1
13
22
Step 6 - Repeat the process to the end of the coefficients
2 |
1
-3
15
-4
-6
8
2
-2
26
44 76
1
-1
13
22
38 84
Step 7. The last number gives the remainder of the division.
Answer: remainder = 84
Step 8 – The numbers on the last row except the last number determine the coefficients of
the quotient which is one degree less than the polynomial.
Quotient: x 4  x3  13x 2  22x  38
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The terms of the polynomial must be written in descending order of degree. If a term is
missing a 0 must be placed on the synthetic division.
Example 2: Divide 3x 4  7x3  3x  18 by x  3
Note: in this case c = -3 because x + 3 = x - - 3
Step 1 - Write the number c and the coefficients of the polynomial on the same line.
-3 |
3
7
0
-3
18
Step 2 - Bring down the first coefficient two spaces and multiply by c. Place the
product under the second coefficient. Repeat the process
-3 |
3
7
0
-3
18
-9
6
-18
63
3 -2
6
-21
81
Step 3 - The last number gives the remainder of the division.
Answer: remainder = 81
Step 4 – The numbers on the last row except the last number determine the coefficients of
the quotient which is one degree less than the polynomial.
Quotient: 3x3  2x 2  6x  21
Example 3: Divide  3x 2  4x 5  17x  10 by x  5
Note: in this case c = -5 because x + 5 = x - - 5
Arrange the polynomial in descending order:  4x 5  3x 2  17x  10 by x  5
Step 1 - Write the number c and the coefficients of the polynomial on the same line.
-5 |
-4
0
0
-3
17
-10
Step 2 - Bring down the first coefficient two spaces and multiply by c. Place the
product under the second coefficient. Repeat the process
-5 |
-4
0
0
-3
17
-10
20 -100
500
-2485 12340
-4 20 -100
497
-2468 12330
Step 3 - The last number gives the remainder of the division.
Answer: remainder = 12330
Step 4 – The numbers on the last row except the last number determine the coefficients of
the quotient which is one degree less than the polynomial.
Quotient:  4x 4  20x3  100x 2  497x  2468
Division Algorithm: If a polynomial P(x) is divided by a nonzero polynomial d(x), then
there is a quotient q(x) and a remainder polynomial r(x) such that P(x) = d(x)q(x) + r(x).
Example 4: Divide 242 by 15.
16
.
15 | 242
P(x) =d(x)q(x) + r(x).
-15
242 = (15)(16) + 2
92
- 90
2
Dividend P(x)
242
Divisor d(x)
15
Quotient q(x)
16
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Remainder r(x)
2
In example 1.
Dividend P(x)
x5  3x 4  15x3  4x 2  6x  8
Divisor
d(x)
x-2
Quotient q(x)
x 4  x3  13x 2  22x  38
Remainder
r(x)
84
x5  3x 4  15x3  4x 2  6x  8 =(x - 2) ( x 4  x3  13x 2  22x  38 ) + 84
In example 2.
Dividend P(x)
4
3
3x  7x  3x  18
Divisor d(x)
x+3
Quotient q(x)
3
2
3x  2x  6x  21
Remainder r(x)
81
3x 4  7x 3  3x  18  ( x  3) ( 3x3  2x 2  6x  21 ) + 81
Factor theorem: If the remainder that results when dividing a polynomial P(x) by x – c is 0,
then x – c is a factor of the polynomial. The other factor is the quotient. The number c is called a
root or zero of the polynomial P(x)
Proof: P(x) = d(x)q(x) + r(x) by the division algorithm principle
Therefore when the divisor is (x- c) and r = 0 , P(x) = (x – c)q(x)
Example 5: Is x – 1 a factor of x 4  15x 3  10x 2  2x  8 ?. If the answer is yes, express
the polynomial as a product of the two factors found.
1 |
1
15
-10
2
-8
1
16
6
8
1
16
6
8
0, yes x – 1 is a factor because r = 0
and the other factor is x3  16x 2  6x  8
Therefore, x 4  15x3  10x 2  2x  8  ( x  1)( x3  16x 2  6x  8)
The Remainder Theorem: If a polynomial P(x) is divided by x - c, then the remainder is
P(c).
Proof:
 P(x) = d(x)q(x) + r(x) by the division algorithm principle
 P(x) = (x – c)q(x) + r(x) when the divisor is x – c
 To find P(c) replace x by c and obtain
 P(c) = (c – c)q(c) + r
 P(c) = (0)q(c) + r
 P(c) = r
Example 6: Use the remainder theorem and synthetic division to find P(-9) if
P( x)  3x 6  20x5  65x 4  15x3  24x 2  26x  170 ?. If the answer is yes, express the
polynomial as a product of the two factors found.
-9 |
3
20
-65 -15
24
-26
170
-27
63
18 -27
27
-9
3
-7
-2
3
-3
1
161
Therefore P(-9) = 161
Check: 3( 9)6  20(9)5  65(9) 4  15(9)3  24(9) 2  26(9)  170 = 161
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