Chabot Mathematics
§6.4 Divide
PolyNomials
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Review § 6.3
MTH 55
 Any QUESTIONS About
• §6.3 → Complex Rational
Expressions
 Any QUESTIONS About HomeWork
• §6.3 → HW-25
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
§6.4 Polynomial Division
 Dividing by a
Monomial
 Dividing by a
BiNomial
 Long Division
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3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Dividing by a Monomial
 To divide a polynomial by a monomial,
divide each term by the monomial.
 EXAMPLE – Divide:
x5 + 24x4 − 12x3 by 6x
 Solution
Chabot College Mathematics
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x5  24 x 4  12 x3 x5 24 x 4 12 x3



6x
6x
6x
6x
1 51 24 41 12 31
 x 
x  x
6
6
6
1 4
 x  4 x3  2 x 2
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  Monomial Division
 Divide: 21a5 b4  14a3b2  7a 2 b  7a 2 b
 Solution:
21a5b4  14a3b2  7a 2b 21a5b 4 14a3b 2 7a 2b



2
2
2
7a b
7a b 7a b 7a 2b
21 5 2 41  14  3 2 21  7 
  a b  a b  
7
7
 7 
 3a 3b 3  2ab  1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Dividing by a Binomial
 For divisors with more than one
term, we use long division, much as
we do in arithmetic.
 Polynomials are written in
descending order and any missing
terms in the dividend are written in,
using 0 (zero) for the coefficients.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Recall Arithmetic Long Division
 Recall Whole-No. Long Division
157
Divide:
12
Divisor
13
12 157
Quotient
12
37
36
1
Quotient •
•
13
Chabot College Mathematics
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Divisor + Remainder
12 +
1
Remainder
= Dividend
= 157
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
 Use an IDENTICAL Long Division
process when dividing by BiNomials or
Larger PolyNomials; e.g.;
Divide 2x³ + 3x² - x + 1 by x + 2
x  2 2x 3  3x 2  x  1
x + 2 is the
divisor
The quotient
will be here.
Chabot College Mathematics
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2x³ + 3x² - x + 1
is the dividend
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
First divide the first term of the dividend, 2x³,
by x (the first term of the divisor).
This gives 2x².
This will be the
first term of
the quotient.
Chabot College Mathematics
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2x 2
x  2 2x 3  3x 2  x  1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
Now multiply
(x+2) by 2x²
and subtract
Chabot College Mathematics
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2x 2
x  2 2x 3  3x 2  x  1
2x 3  4x 2
x 2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
2x 2
x  2 2x 3  3x 2  x  1
Bring down the
next term, -x.
Chabot College Mathematics
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2x 3  4x 2
x 2 x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
Now divide –x²,
2x 2 x
the first term of x  2 2x 3  3x 2  x  1
–x² - x, by x, the
3
2
2
x

4
x
first term of the
divisor
x 2 x
which gives –x.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
2x 2 x
x  2 2x 3  3x 2  x  1
2x 3  4x 2
Multiply (x +2) by -x
and subtract
Chabot College Mathematics
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x 2 x
x 2  2x
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
2x 2 x
x  2 2x 3  3x 2  x  1
2x 3  4x 2
Bring down the
next term, 1
Chabot College Mathematics
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x 2 x
x 2  2x
x 1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
2x 2 x 1
x  2 2x 3  3x 2  x  1
Divide x, the first
term of x + 1, by x,
the first term of
the divisor
which gives 1
Chabot College Mathematics
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2x 3  4x 2
x 2 x
x 2  2x
x 1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
2x 2 x 1
x  2 2x 3  3x 2  x  1
2x 3  4x 2
x 2 x
x 2  2x
Multiply x + 2 by 1
and subtract
Chabot College Mathematics
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x 1
x 2
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Binomial Div.  Step by Step
2x 2 x 1
x  2 2x 3  3x 2  x  1
The quotient is
2x² - x + 1
2x 3  4x 2
x 2 x
x 2  2x
The remainder is –1.
x 1
x 2
1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  BiNomial Division
 Divide x2 + 7x + 12 by x + 3.
 Solution
x
x  3 x 2  7 x  12
( x  3 x )
2
4x
Chabot College Mathematics
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Multiply (x + 3) by x, using
the distributive law
Subtract by changing signs
and adding
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  BiNomial Division
 Solution – Cont.
x+4
2
x  3 x  7 x  12
( x 2  3 x )
4 x  12
 ( 4 x  12 )
0
Chabot College Mathematics
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Bring Down the +12
Multiply 4 by the divisor, x + 3,
using the distributive law
Subtract
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  BiNomial Division
 Divide 15x2 − 22x + 14 by (3x − 2)
5x  4
 Solution
2
3x  2 15x  22 x  14
(15 x 2  10 x)
 12 x  14
(12 x  8)
 The answer is 5x − 4
with R6. We can also
write the answer as:
Chabot College Mathematics
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6
6
5x  4 
.
3x  2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  BiNomial Division
 Divide x5 − 3x4 − 4x2 + 10x by (x − 3)
4
 4x  2
x
 Solution
2
3
4
5
x 3
x  3 x  0 x  4 x  10 x  0
( x 5  3 x 4 )
0 x 3  4 x 2  10 x
  4 x 2  12 x 
 The
Result
Chabot College Mathematics
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6
x  4x  2 
.
x 3
4
 2x  0
( 2 x  6)
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Divide 3x2 − 4x − 15 by x − 3
 SOLUTION: Place the TriNomial under
the Long Division Sign and start the
Reduction Process
3x
x  3 3 x  4 x  15
2
(3x 2  9 x)
5x 15
Chabot College Mathematics
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Divide 3x2 by x:
3x2/x = 3x.
Multiply x – 3 by 3x.
Subtract by mentally changing
signs and adding −4x + 9x = 5x.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Divide 3x2 − 4x − 15 by x − 3
 SOLUTION: next divide the leading
term of this remainder, 5x, by the
leading term of the divisor, x.
3x  5
x  3 3 x 2  4 x  15
(3x 2  9 x)
Divide 5x by x: 5x/x = 5.
5x 15
(5 x  15)
0
Multiply x – 3 by 5.
Subtract. Our remainder is now 0.
 CHECK: (x − 3)(3x + 5) = 3x2 − 4x − 15
 The quotient is 3x + 5.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Formal Division Algorithm
 If a polynomial F(x) is divided by a
polynomial D(x), with D(x) ≠ 0, there are
unique polynomials Q(x) and R(x) such that
F(x)
Dividend
= D(x)
Divisor
• Q(x) + R(x)
Quotient
Remainder
 Either R(x) is the zero polynomial, or
the degree of R(x) is LESS than the
degree of D(x).
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
PolyNomial Long Division
1. Write the terms in the dividend and the
divisor in descending powers of the
variable.
2. Insert terms with zero coefficients in
the dividend for any missing powers of
the variable
3. Divide the first terms in the dividend by
the first terms in the divisor to obtain
the first term in the quotient.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
PolyNomial Long Division
4. Multiply the divisor by the first term in
the quotient, and subtract the product
from the dividend.
5. Treat the remainder obtained in Step 4
as a new dividend, and repeat Steps 3
and 4. Continue this process until a
remainder is obtained that is of lower
degree than the divisor.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  TriNomial Division
 Divide x 4  13x 2  x  35 by x 2  x  6.
 SOLN
x2  x
x 2  x  6 x 4  0x 3  13x 2  x  35
x 
4
x  6x
3
2
x  7x  x  35
3
2
x3 
x 2  6x
6x  7x  35
2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  TriNomial Division
x2  x  6
 SOLN
2
4
3
2
cont. x  x  6 x  0x  13x  x  35
4
3
2
x  x  6x
x 3  7x 2  x  35
x3 
x 2  6x
6x  7x  35
2
6x  6x  36
x 1
2
Chabot College Mathematics
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Example  TriNomial Division
 Divide x 4  13x 2  x  35 by x 2  x  6.
 The Quotient = x  x  6.
 The Remainder = x  1.
2
 Write the Result in Concise form:
x 4  13x 2  x  35
x 1
2
 x x6 2
.
2
x x6
x x6
Chabot College Mathematics
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
WhiteBoard Work
 Problems From §6.4 Exercise Set
• 30, 32, 40

BiNomial
Division
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
All Done for Today
Polynomial
Division
in base2
From UC Berkeley Electrical-Engineering 122 Course
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
4
5
6
y
5
2
y
1
x
4
0
-6
-5
-4
-3
-2
-1
3
0
1
2
3
4
-1
-2
2
-3
1
-4
x
-5
5
-6
0
-3
-2
-1
0
1
2
3
-1
4
-7
-8
-2
-9
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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file =XY_Plot_0211.xls
-10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt
5
6