Download Division of Polynomials

Section 2-3
Long and Synthetic
Division of Polynomials
Objectives
• I can use long division to
divide two polynomials
• I can use synthetic division
to divide a polynomial by a
binomial (x – r)
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Dividing Numbers
Quotient
Divisor
4
4 16
Dividend
When you divide a number
by another number and
there is no remainder:
Then the divisor is a
factor!!
Also the quotient becomes
another factor!!!
3
Dividing Polynomials
Long division of polynomials is similar to long division of
whole numbers.
When you divide two polynomials you can check the answer
using the following:
dividend = (quotient • divisor) + remainder
The result is written in the form:
remainder
dividend  divisor  quotient +
divisor
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Dividing Polynomials
Example: Divide x2 + 3x – 2 by x + 1 and check the answer.
x + 2
x  1 x 2  3x  2
x2 +
x
2x – 2
2x + 2
–4
remainder
Answer: x + 2 +
x2
1. x x 
x
x
2. x ( x  1)  x 2  x
2
3. ( x 2  3 x )  ( x 2  x )  2 x
2x
4. x 2 x 
2
x
5. 2 ( x  1)  2 x  2
6. ( 2 x  2 )  ( 2 x  2 )   4
–4
x 1
5
Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer.
x2 + x + 3
2x  2 2x  0x  4x  1
3
2
2x3 – 2x2
2x2 + 4x
2x2 – 2x
6x – 1
6x – 6
5
Answer: x2 + x + 3 
5
2x  2
Write the terms of the dividend in
descending order.
Since there is no x2 term in the
dividend, add 0x2 as a placeholder.
3
2
x
1.
2. x 2 ( 2 x  2 )  2 x 3  2 x 2
 x2
2x
2x2
3
3
2
2
3. 2 x  ( 2 x  2 x )  2 x
4.
x
2x
5. x ( 2 x  2 )  2 x 2  2 x
6. ( 2 x 2  4 x )  ( 2 x 2  2 x )  6 x
8. 3 ( 2 x  2 )  6 x  6
7. 6 x  3
2x
9. ( 6 x  1)  ( 6 x  6 )  5  remainder
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Example: Divide x2 – 5x + 6 by x – 2.
x – 3
x  2 x2  5x  6
x2 – 2x
– 3x + 6
– 3x + 6
0
Answer: x – 3 with no remainder.
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Dividing by Synthetic Division
• Synthetic Division is a method to divide any
polynomial by a binomial.
• The steps must be followed exactly in order
or you will not get the correct end result
• The following slide shows the steps for one
complete problem.
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Find: (6x3- 19x2 + x + 6)  (x-3)
•
•
•
•
•
Step 1: Rewrite the dividend with
all terms. If a term is missing,
insert a zero for that term.
Bring down the coefficients from
the dividend and make a row.
Next identify the divisor. It must be
in the format (x-r). Bring down r
and put in a box on the left. Draw a
line.
Bring down 1st coefficient under
the line. Multiply it by “r” and add
to next column. Then repeat.
New row of numbers are the
coefficients of the quotient starting
with one power less.
•
6x3 – 19x2 + 1x + 6
6
3
6
-19
1
6
18
-3
-6
-1
-2
0
6x2 – 1x – 2 (No remainder)
9
Find: (4x4- 5x2 + 2x + 4)  (x+1)
•
•
•
•
•
Step 1: Rewrite the dividend with
all terms. If a term is missing,
insert a zero for that term.
Bring down the coefficients from
the dividend and make a row.
Next identify the divisor. It must be
in the format (x-r). Bring down r
and put in a box on the left. Draw a
line.
Bring down 1st coefficient under
the line. Multiply it by “r” and add
to next column. Then repeat.
New row of numbers are the
coefficients of the quotient starting
with one power less.
• 4x4 + 0x3 – 5x2 + 2x + 4
4
-1
4
0
-5
2
4
-4
4
1
-3
-4
-1
3
1
1
4 x  4 x  1x  3 
x 1
3
2
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Synthetic division is a shorter method of dividing polynomials.
This method can be used only when the divisor is of the form
x – a. It uses the coefficients of each term in the dividend.
Example: Divide 3x2 + 2x – 1 by x – 2 using synthetic division.
Since the divisor is x – 2, a = 2.
coefficients of the dividend
value of a
2
3
3
2
–1
6
16
8
15
coefficients of quotient
Answer: 3x + 8 
1. Bring down 3
2. (2 • 3) = 6
3. (2 + 6) = 8
4. (2 • 8) = 16
5. (–1 + 16) = 15
remainder
15
x2
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Homework
• WS 4-1
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