Transcript Introduction to Algebra

Polynomial Division
Division
Long and Short
Polynomial Functions

Division of Polynomials

Long Division (by hand):
40
Quotient
245 ) 9815
980
Remainder
15
So,
9815
15
3
245 = 40 + 245 = 40 + 49
 Can we do this with polynomials ?
We just did ! *
* See Notes page
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Polynomial Division
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Polynomial Functions

General Division of Polynomials

For polynomials f(x) and d(x) , d(x) ≠ 0
f(x)
r(x)
Q(x)
+ d(x)
=
d(x)

where Q(x) is the quotient , r(x) is the
remainder , d(x) is the divisor and f(x)
is the dividend
Thus
f(x) = d(x) ∙ Q(x) + r(x)
where either r(x) = 0 or deg r(x) < deg d(x)
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Polynomial Functions

Algebraic Monomial Division

4 + 27x3 – 9x2 + 6x – 2
12x
Example:
3x2
27x3 – 9x2
12x4
6x – 2
+ 3x2
= 3x2 + 3x2
3x2
3x2
= 4x2 + 9x – 3 + 2x – 2 2
3x
=
4x2
6x – 2
–
+ 9x 3 + 3x2
Remainder
Quotient
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Polynomial Functions

Arithmetic Monomial Division

Example
Quotient
4x2 + 9x – 3
3x2 ) 12x4 + 27x3 – 9x2 + 6x – 2
12x4
27x3
27x3
Remainder
– 9x2
– 9x2
6x – 2
Note: deg r(x) = 1 < deg d(x) = 2
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Polynomial Functions

Division by Linear Binomials

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Example
Quotient
x3 + x2 – 2x
x + 2 ) x4 + 3x3
– 4x + 1
x4 + 2x3
x3
x3 + 2x2
Remainder
– 2x2 – 4x
– 2x2 – 4x
1
Question:
Can we do this faster or more simply ?
Polynomial Division
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Synthetic Division
 Example:

2x3 – 7x2 +19x – 34
x + 2 ) 2x4 – 3x3 + 5x2 + 4x + 3
2x4 + 4x3
–7x3 + 5x2
3 – 14x2
–7x
Arithmetic
2 + 4x
19x
operations
19x2 + 38x
involve only the
– 34x + 3
coefficients
– 34x – 68
71
Remainder
NOTE: deg r(x) = 0 < 1 = deg d(x)
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Synthetic Division
 Example:
x + 2 ) 2x4 – 3x3 + 5x2 + 4x + 3
 Using synthetic division we deal
only with the coefficients
d(x) = x + 2
Subtract
d(x) = x – (–2)
Add
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2 2 –3 5 4 3
4 –14 38 –68
2 –7 19 –34 71
–2 2 –3 5 4 3
–4 14 –38 68
2 –7 19 –34 71
Polynomial Division
Remainder
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Polynomial Functions

Degree Facts

For any polynomials A(x), B(x)
deg (A(x) • B(x)) = deg A(x) + deg B(x)
deg (A(x) + B(x)) = max { deg A(x), deg B(x) }

Examples:
2
3
2
 deg ( (x + 1) • (3x – 4x + 5x + 7) )
=2+3=5

deg ( (3x2 – 4) + (2x4 + 6x + 3) )
= max { 2, 4 } = 4
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Polynomial Functions
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Division Algorithm for Polynomials
Consider polynomial functions f(x), d(x) with
0 < deg d(x) < deg f(x)
There exist unique polynomial functions
Q(x) and r(x)
such that
f(x) = d(x) • Q(x) + r(x)
where either r(x) = 0 or deg r(x) < deg d(x)
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Polynomial Functions

Division Algorithm (continued)
f(x) = d(x) • Q(x) + r(x)
with r(x) = 0 or deg r(x) < deg d(x)
Note: This just says that
f(x)
r(x)
Q(x) + d(x)
d(x) =
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Polynomial Functions
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Division Algorithm (continued)
f(x)
r(x)
Q(x) + d(x)
d(x) =
Example:
x2 – 1
=x+1
x–1
deg f(x) = 2
deg d(x) = 1 , r(x) = 0
So,
f(x) = d(x) • Q(x) + r(x)
becomes
x2 – 1 = (x – 1) • (x + 1) + 0
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Polynomial Functions

Division Algorithm and Degrees
Given
f(x) = d(x) • Q(x) + r(x)
where
either r(x) = 0 or deg r(x) < deg d(x)
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Polynomial Functions
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Division Algorithm and Degrees
Given
f(x) = d(x) • Q(x) + r(x)
Question: Suppose
deg d(x) = m > 0 ,
deg Q(x) = n ,
deg f(x) = p > m
What is the relationship among m, n and p ?
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Polynomial Functions

Division Algorithm and Degrees
f(x) = d(x) • Q(x) + r(x)
p = deg f(x)
= deg ( d(x) • Q(x) + r(x) )
= max { deg (d(x) • Q(x)) , deg r(x) }
= max { (deg d(x) + deg Q(x)) , deg r(x) }
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Polynomial Functions
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Division Algorithm and Degrees
p = deg f(x)
= max { (deg d(x) + deg Q(x)) , deg r(x) }
= max { (m + n) , deg r(x) }
Since deg r(x) < deg d(x)
=m
<m+n
then max { (m + n) , deg r(x) } = m + n
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Polynomial Functions

Division Algorithm and Degrees
f(x) = d(x) • Q(x) + r(x)
p = deg f(x)
= max { (m + n) , deg r(x) }
=m+n
So
p=m+n
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Think about it !
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