3.3B Dividing Polynomials
Objectives:
A.APR.2: Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the
remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x).
A.APR.6: Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form
q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x)
less than the degree of b(x).
For the Board:
You will be able to use long division and synthetic division to divide polynomials.
Anticipatory Set:
Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only
the coefficients of the polynomials involved.
For synthetic division the polynomial must be written in perfect standard form and the divisor must be
in the form (x – a).
Instruction:
Open the book to page 167 and read about synthetic division.
Example: Divide (2x2 + 7x + 9)  (x + 2) by using synthetic division.
Step 1: Write the value of a for the division (x – a)
using the separator .
Step 2: Write the coefficients only of the dividend
in perfect descending order after the separator.
Step 3: Leave a blank row. Draw a line under the blank row.
Step 4: Bring down below the line the first coefficient.
Step 5: Multiply the a by this coefficient and place the
answer above the line under the next coefficient.
Step 6: Add the two numbers and place the answer below
the line.
Step 7: Repeat steps 5 and 6 till complete.
Step 8: Place a separator around the last term.
1, 2, 3.
-2| 2
7
9
_________
4, 5, 6. - 2 | 2
7
9
4
7, 8.
2
3
-2| 2
7
9
4 6
2
3 |3
The quotient is represented by the numbers below the horizontal bar. The boxed number is the
remainder. The quotient always has degree 1 less than the degree of the original dividend.
Dividend: 2x2 + 7x + 9
Quotient: 2x + 3
Remainder: 3
3
2x2 + 7x + 9 = (2x + 3)(x + 2) + 3
or
(2x2 + 7x + 9)  (x + 2) = 2x  3 
x2
Open the book to page 167 and read example 2.
Example: Divide using synthetic division.
a. (3x2 - 8x – 2)  (x – 3)
3| 3 8  2
9
3
3 1 | 1
1
or 3x2 - 8x – 2 = (x – 3)(3x + 1) + 1
x2
b. (3x4 – x3 + 5x – 1)  (x + 2)
 2 | 3 1 0
5 1
3x  1 
 6 14  28 46
3  7 14  23 | 45
45
or
3x 3  7x 2  14x - 23 
x2
3x4 – x3 + 5x – 1 = (x + 2)( 3x 3  7x 2  14x - 23 ) + 45
White Board Activity:
Practice: Divide by using synthetic division.
a. (6x2 – 5x – 6)  (x – 3)
3| 6 5 6
b. (x2 + 3x – 18)  (x + 6)
 6 | 1 3  18
6
18 39
6
13 | 33
33
6x  13 
x3
6x2 – 5x – 6 = (x – 3)(6x + 13) + 33
1
3
|0
x – 3 (no remainder)
x2 + 3x – 18 = (x + 6)(x – 3)
Synthetic division can be used to evaluate polynomials.
In this case it is called synthetic substitution.
The process is the same, the interpretation of the results is different.
Example: Use synthetic substitution to find P(3) when P(x) = x3 – 4x2 + 5x + 1.
3| 1 4
5 1
3 3 6
1 1
P(3) = 17
18
2 |7
This is called the Remainder Theorem
Remainder Theorem
If the polynomial function P(x) is divided by x – a, then the remainder r is P(a).
P(a) = r when r is the remainder of P(x)  (x – a).
Open the book to page 168 - 169 and read example 3.
Example: Use synthetic substitution to evaluate the polynomial for the given value.
a. P(x) = 2x3 + 5x2 – x + 7 for x = 2
2| 2
5 1 7
4 18 34
2 9 17 | 41
P(2) = 41
b. P(x) = 6x4 – 25x3 – 3x + 5 for x = -1/3
 1 / 3 | 6  25
0 3 5
 23
9 3 2
6  27
P(-1/3) = 7
9
6|7
White Board Activity:
Practice: Use synthetic substitution to evaluate the polynomial for the given value.
a. P(x) = x3 + 3x2 + 4 for x = -3
b. P(x) = 5x2 + 9x + 3 for x = 1/5
1/ 5 | 5
9 3
3| 1 3 0 4
3 0
1
0
1 2
0
5
0 | 4
P(-3) = 4
10 | 5
P(1/5) = 5
Open the book to page 169 and read example 4.
Example: Write an expression that represents the area of the top face of a rectangular prism when the
height is x + 2 and the volume of the prism is x3 – x2 – 6x.
V = Bh
x3 – x2 – 6x = B(x + 2)
B = (x3 – x2 – 6x)  (x + 2)
 2 | 1 1  6
2
1
3
6
|0
B = x2 – 3x
White Board Activity:
Practice: Write an expression for the length of a rectangle with width y – 9 and area y2 – 14y + 45.
A = LW
y2 – 14y + 45 = L(y – 9)
L = (y2 – 14y + 45)  (y – 9)
9| 1
 14
45
9  45
1
L=y–5
5
|0
Assessment:
Question student pairs.
Independent Practice:
Text: pgs. 170 – 171 prob. 5 – 11, 19 – 28, 40, 44, 45, 47, 48.
For a Grade:
Text: pgs. 170 – 171 prob. 6, 10, 20, 28.