Download Chapter 2, Section 4

Chapter 2
Polynomial and
Rational Functions
2.4 Dividing Polynomials:
Remainder and Factor
Theorems
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
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Objectives:
•
•
•
•
Use long division to divide polynomials.
Use synthetic division to divide polynomials.
Evaluate a polynomial using the Remainder Theorem.
Use the Factor Theorem to solve a polynomial
equation.
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Divide a polynomial by
a) monomial
b)common factor
2x  6x  4x
2x
3
2
3
2
x  2x  8
x2
2
2x 6x
4x


2x 2x 2x
( x  2)( x  4)
x2
x 2  3x  2
x4
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Example: Divide (2x2 + 3x – 4) ÷ (x – 2)
divisor
dividend
Is 1) Think,
+ 3x
– many
4) factorable
?
how
times
(2x2
does x go into 2x2 ?
No
2) Multiply by the divisor.
Rewrite in long
2x + 7
(x – 2)
division form...
2
2x + 3x – 4
(2x2 – 4x)
7x – 4
3) Subtract and bring
down
Then
we4. have to use other(7x
methods
– 14)
4) Think, how many times
of division - long division
does x go into 7x ?
10
remainder
5) Multiply by the divisor.
6) Subtract.
Write the result like this...
quotient
10
2x  7 
x2
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Dividing Polynomials - Vocabulary
Long division of polynomials is similar to long
division of whole numbers.
The result is written in the form:
dividend  divisor  quotient +
remainder
divisor
When you divide two polynomials you can check the
answer using the following:
dividend = (quotient • divisor) + remainder
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Try This: Long Division of Polynomials
3
2
2
x

3
x
 11x  7 by x  3
Divide
22x
x 22  3x  2
x  3 2 x3  3x 2  11x  7
(22x3  6 x 2 )
3x 2  11x
2
(33x  9 x)
2 x  7
(22 x  6)
6
1
The quotient is
1
2
2 x  3x  2 
.
x 3
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IMPORTANT to remember
If either dividend or divisor is missing term/s with x
insert that term/s with coefficient of 0
as a place-
holder and then divide. This way all terms will be
aligned as you carry out your division.
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(3
x
 2 x  5)  (2 x  3)
Ex:
(3x3  0x
0 x2  2 x  5)  (2 x  3)
(2 x 4  6)  (4 x 2  1)
or4
(2 x  0x
0 x3  0x
0 x22  00x
x  6)  (4 x 2  0x
0 x  1)
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Try This: Long Division (p3 – 6) ÷ (p – 1)
2
(p – 1)
p + p +1
3
2
p + 0p + 0p – 6
3
Be sure to add “place-holders”
p – p2
p2 + 0p
p2 – p
for missing terms...
5
p  p 1
p 1
2
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p–6
p–1
–5
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Try This: Long Division (x3 –2x2 +x-3) ÷ (x – 4)
(x – 4)
3
2
x – 2x + x – 3
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Synthetic division
used ONLY when the divisor is in the form (x – k).
Ex: Use synthetic division for the following:
3
2
(2x – 7x – 8x + 16) ÷ (x – 4)
1) Check for missing terms - None 
2) Write down the coefficients in descending order,
and k of the divisor in the form x – k :
These are the
coefficients of the
4
k
2 –7 –8 16
quotient (and the
1)Bring down
8 4 -16
remainder is 0)
the first
coefficient.
2 1 –4 0
2) Multiply
this by k
3) Add the
column.
 2x  x  4
2
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You Try: Synthetic division
(3x³ -4x² +2x -1) / (x-1)
1
3
-4
2
-1
3
-1
1
3
-1
1
3x2
-1x + 1
0
These are the
coefficients of the
quotient (and the
remainder again is 0)
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Try again: Divide (5x3 + x2 – 7) ÷ (x + 1)
We need a place-holder
Notice that
k is –1 since
–1
synthetic division
works for divisors
5
1
–5
5 –4
0
–7
4 –4
4 –11
in the form (x – k).
11
5x  4 x  4 
x 1
2
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FOLDABLE
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Long Division of Polynomials
1. Arrange the terms of both the dividend and the
divisor in descending powers of any variable.
2. Divide the first term in the dividend by the first term
in the divisor. The result is the first term of the
quotient.
3. Multiply every term in the divisor by the first term
in the quotient. Write the resulting product beneath
the dividend with like terms lined up.
4. Subtract the product from the dividend.
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Long Division of Polynomials (continued)
5. Bring down the next term in the original dividend
and write it next to the remainder to form a new
dividend.
6. Use this new expression as the dividend and repeat
this process until the remainder can no longer be
divided. This will occur when the degree of the
remainder (the highest exponent on a variable in the
remainder) is less than the degree of the divisor.
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Synthetic Division
1. Arrange the polynomial in descending powers, with
a 0 coefficient for any missing term.
2. Write c for the divisor, x – c. To the right, write the
coefficients of the dividend.
3. Write the leading coefficient of the dividend on the
bottom row.
4. Multiply c times the value just written on the bottom
row. Write the product in the next column in the
second row.
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Synthetic Division (continued)
5. Add the values in this new column, writing the sum
in the bottom row.
6. Repeat this series of multiplications and additions
until all columns are filled in.
7. Use the numbers in the last row to write the quotient,
plus the remainder above the divisor. The degree of
the first term of the quotient is one less than the
degree of the first term of the dividend. The final
value in this row is the remainder.
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Example: Using Synthetic Division
Use synthetic division to divide x  7 x  6 by x + 2.
The divisor must be in form x – c. Thus, we write x + 2
as x – (–2). This means that c = –2. Writing a 0
coefficient for the missing x2 term in the dividend, we
can express the division as follows:
3
x  (2) x3  0 x 2  7 x  6
Now we are ready to perform the synthetic division.
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Synthetic Division (continued)
Use synthetic division to divide x  7 x  6 by x + 2.
3
2 1
0 7 6
22 44 6
11 22 3 0
The quotient is x 2  2 x  3.
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The Remainder Theorem
If the polynomial f(x) is divided by x – c, then the
remainder is f(x).
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Example: Using the Remainder Theorem to Evaluate a
Polynomial Function
Given f ( x)  3x  4 x  5 x  3 use the Remainder
Theorem to find f(–4).
We use synthetic division to divide.
3
2
4 3 4  5  3
12
 
12 32 108
3 8 27 105
The remainder, –105, is the value of f(–4). Thus,
f(–4) = –105.
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The Factor Theorem
Let f(x) be a polynomial.
a. If f(x) = 0, then x – c is a factor of f(x).
b. If x – c is a factor of f(x), then f(c) = 0.
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Example: Using the Factor Theorem
Solve the equation 15 x3  14 x 2  3x  2  0 given that
–1 is a zero of f ( x)  15 x3  14 x 2  3x  2.
We are given that –1 is a zero of
f ( x)  15 x3  14 x 2  3x  2. This means that f(–1) = 0.
Because f(–1) = 0, the Factor Theorem tells us that x + 1
is a factor of f(x). We’ll use synthetic division to divide
f(x) by x + 1.
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Example: Using the Factor Theorem
Solve the equation 15 x3  14 x 2  3x  2  0 given that
–1 is a zero of f ( x)  15 x3  14 x 2  3x  2.
We’ll use synthetic division to divide f(x) by x + 1.
1 15 14 3 2
 15 1 2
15 1 2 0
(15 x  14 x  3x  2)  ( x  1)  15 x  x  2
This means that
(15 x3  14 x 2  3x  2)  ( x  1)(15 x 2  x  2).
3
2
2
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Example: Using the Factor Theorem
(continued)
Solve the equation 15 x3  14 x 2  3x  2  0 given that
–1 is a zero of f ( x)  15 x3  14 x 2  3x  2.
(15 x3  14 x 2  3x  2)  ( x  1)(15 x 2  x  2)
(15 x3  14 x 2  3x  2)  ( x  1)(3x  1)(5 x  2)
x  1  0  x  1
1
3x  1  0  x  
3
2
5x  2  0  x 
5
The solution set is
1 2
1,  , .
3 5

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
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