[email protected] mth55_lec-33_sec_6-5_synthetic_division.ppt 1 bruce mayer, pe chabot...
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[email protected] • MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
§6.5 Synthetic§6.5 SyntheticDivisionDivision
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Bruce Mayer, PE Chabot College Mathematics
Review §Review §
Any QUESTIONS About• §6.4 → PolyNomial Long Division
Any QUESTIONS About HomeWork• §6.4 → HW-26
6.4 MTH 55
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Bruce Mayer, PE Chabot College Mathematics
StreamLining Long DivisionStreamLining Long Division To divide a polynomial by a binomial of
the type x − c, we can streamline the usual procedure to develop a process called synthetic division.
Compare the following. In each stage, we attempt to write a bit less than in the previous stage, while retaining enough essentials to solve the problem. At the end, we will return to the usual polynomial notation.
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Bruce Mayer, PE Chabot College Mathematics
Stage 1: Synthetic DivisionStage 1: Synthetic Division
When a polynomial is written in descending order, the coefficients provide the essential information.
3 21 2 0 6 12x x x x
22 2 4x x
3 22 2x x22 6x x 2 22x x
4 12x 44x
16
1 1 2 0 6 12 2 2 4
2 22 6
22 4 12
44 16
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Bruce Mayer, PE Chabot College Mathematics
Leading Coefficient ImportanceLeading Coefficient Importance
Because the leading coefficient in the divisor is 1, each time we multiply the divisor by a term in the answer, the leading coefficient of that product duplicates a coefficient in the answer. In the next stage, we don’t bother to duplicate these numbers. We also show where the other +1 is used and drop the first 1 from the divisor.
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Bruce Mayer, PE Chabot College Mathematics
Stage 2: Synthetic DivisionStage 2: Synthetic Division
3 21 2 0 6 12x x x x
22 2 4x x
3 22 2x x22 6x x 22 2x x
4 12x 4 4x
16
2 0 121 6 2 2 4
22 6
24 12
416
MultiplySubtractMultiply
Multiply
Subtract
Subtract
To simplify further, we now reverse the sign of the 1 in the divisor and, in exchange, add at each step in the long division.
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Bruce Mayer, PE Chabot College Mathematics
Stage 3: Synthetic DivisionStage 3: Synthetic Division
The blue numbers can be eliminated if we look at the red numbers instead.
Replace the 1 with -13 21 2 0 6 12x x x x
22 2 4x x
3 22 2x x22 6x x 2 22x x
4 12x 44x
16
61 2 20 1 2 2 4
262 2
14 2 4
16
MultiplyAddMultiply
MultiplyAdd
Add
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Bruce Mayer, PE Chabot College Mathematics
Stage 4: Synthetic DivisionStage 4: Synthetic Division Don’t lose sight
of how the products −2, 2, and 4 are found. Also, note that the −2 and −4 preceding the remainder 16 coincide with the −2 and −4 following the 2 on the top line. By writing a 2 to the left of the −2 on the bottom line, we can eliminate the top line in stage 4 and read our answer from the bottom line. This final stage is commonly called synthetic division.
3 21 2 0 6 12x x x x
22 2 4x x
3 22 2x x22 6x x 2 22x x
4 12x 44x
16
1 2 0 6 12
2 2 4
2 2 42 4 16
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Bruce Mayer, PE Chabot College Mathematics
Stage 5: Synthetic DivisionStage 5: Synthetic Division
For (2x2 − 2x − 4)(x + 1) then:• quotient is 2x2 − 2x − 4.
• remainder is 16.
1 2 0 6 12
2 2 4
2 2 42 4 16
1 2 0 6 12
2 2 4
2 2 4 16 This is the remainder.
This is the zero-degree coefficient.
This is the first-degree coefficient.
This is the second-degree coefficient.
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Bruce Mayer, PE Chabot College Mathematics
Form of the Divisor → Form of the Divisor → xx − − cc
Remember that in order for the Synthetic Division method to work, the divisor must be of the form x – c, that is, a variable minus a constant
Both the coefficient and the exponent of the variable MUST be 1
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Bruce Mayer, PE Chabot College Mathematics
Synthetic DivisionSynthetic Division
1. Arrange the coefficients of F(x) in order of descending powers of x, supplying zero as the coefficient of each missing power.
2. Replace the divisor x − c with c.
3. Bring the first (leftmost) coefficient down below the line. Multiply it by c, and write the resulting product one column to the right and above the line.
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Bruce Mayer, PE Chabot College Mathematics
Synthetic DivisionSynthetic Division
4. Add the product obtained in Step 3 to the coefficient directly above it, and write the resulting sum directly below it and below the line. This sum is the “newest” number below the line.
5. Multiply the newest number below the line by c, write the resulting product one column to the right and above the line, and repeat Step 4
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Bruce Mayer, PE Chabot College Mathematics
Example Example Synthetic Division Synthetic Division
Use synthetic division to divide
2x4 x3 16x2 18 by x 2.
2 2 1 16 0 18
4 6 20 40
2 3 10 20 22
SOLUTIONby SyntheticDivision
Thus the Result• Quotient → 2x3 3x2 10x 20
• Remainder → −22
Or 2x3 3x2 10x 20
22
x 2.
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Bruce Mayer, PE Chabot College Mathematics
Example Example Synthetic Division Synthetic Division
Use synthetic division to divide (x4 − 9x3 − 7x2 + 10)/(x + 5)
SOLUTION: The divisor is x + 5, so write −5 at the left
−5 1 9 7 0 10
5 70 315 1575
1 14 63 315 1585
Thus theAlgebra ANS
3 2 158514 63 315 .
5x x x
x
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Bruce Mayer, PE Chabot College Mathematics
Long vs. SyntheticLong vs. Synthetic
compare the bare essentials for finding the quotient for the Two Division Methods
2 7 14
5
x x
x
1 7 14
5
1 2
−52
5 7 14x x x
2 5x x
x
2 14x 2 10x
4
2
Long Division Synthetic Division
Coefficients of the polynomial
104
c in the divisor, x - c
remaindercoefficients of the quotient,
x+2
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Bruce Mayer, PE Chabot College Mathematics
Example Example Synthetic Division Synthetic Division
Use synthetic division to divide
(2x3 − x2 − 43x + 60) ÷ (x − 4). SOLUTION by Synthetic Division
4 2 1 43 60
2
4 2 1 43 60 8
2 7
Write the 4 of x – 4 and the coefficients of the dividend.
Bring down the first coefficient.
Multiply 2 by 4 to get 8.
Add −1 and 8.
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Bruce Mayer, PE Chabot College Mathematics
Example Example Synthetic Division Synthetic Division
SOLUTION by Synthetic Division
The answer is 2x2 + 7x − 15 with R 0, or just 2x2 + 7x − 15
4 2 1 43 60 8 28
2 7 15Multiply 7 by 4 to get 28.
Add −43 and 28.
4 2 1 43 60 8 28 60
2 7 015
Multiply −15 by 4 to get -60.
Add 60 and −60.
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Bruce Mayer, PE Chabot College Mathematics
Remainder TheoremRemainder Theorem Because the remainder is 0, the last example
shows that x – 4 is a factor of 2x3 – x2 – 43x + 60 and that we can write 2x3 – x2 – 43x + 60 as (x – 4)(2x2 + 7x – 15). Using this result and the principle of zero products, we know that if f(x) = 2x3 – x2 – 43x + 60, then f(4) = 0.
In this example, the remainder from the division, 0, can serve as a function value. Remarkably, this pattern extends to nonzero remainders. The fact that the remainder and the function value coincide is predicted by the remainder theorem.
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Bruce Mayer, PE Chabot College Mathematics
Remainder TheoremRemainder Theorem
The remainder obtained by dividing a PolyNomial f(x) by
(x − c) is f(c). In other words, If a number c is
substituted for x in a polynomial f(x), then the result, f(c), is the remainder that would be obtained by dividing f(x) by x − c. That is, if f(x) = (x − c) • Q(x) + R, then f(c) = R.
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Bruce Mayer, PE Chabot College Mathematics
Example Example Find Remainder Find Remainder
Find the remainder when dividing the following polynomial by (x − 1)
F x 2x5 4x3 5x2 7x 2
SOLUTION: By the Remainder Theorem, F(1) is the remainder
F 1 2 1 5 4 1 3 5 1 2 7 1 2
2 4 5 7 2 2
The Remainder is −2
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Bruce Mayer, PE Chabot College Mathematics
Example Example Evaluate PolyNomial Evaluate PolyNomial
Find for f(−3) for PolyNomial
f x x4 3x3 5x2 8x 75.
Evaluate by straight Substitution
Use The Remainder Theorem• If we know R for f(x)(x − [−3]), then f(−3) is
simply R → find R by Synthetic Division
f 3 3 4 3 3 3 5 3 2 8 3 75 6
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Bruce Mayer, PE Chabot College Mathematics
Example Example Evaluate PolyNomial Evaluate PolyNomial
Find for f(−3) for PolyNomial
f x x4 3x3 5x2 8x 75.
SyntheticDivision:(x+3) divisor
Then by the Remainder Theorem
f(−3) = R = 6• Same Result as by Direct Substitution
3 1 3 5 8 75
3 0 15 69
1 0 5 23 6
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Bruce Mayer, PE Chabot College Mathematics
Factor TheoremFactor Theorem
A polynomial f(x) has (x – c) as a factor if and only if f(c) = 0
In other words, If f(x) is divided by (x − c) and the remainder is 0, then f(c) = 0. This means the c is solution of the equation f(x) = 0• Can use this to VERIFY potential Solns
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Bruce Mayer, PE Chabot College Mathematics
Example Example Factor Theorem Factor Theorem
Show that x = 2 is a solution to
3x2 2x2 19x 6 0.
Then find the remaining solutions to this polynomial equation
SOLUTION: If 2 is a solution, f(2) = 0 then (x – 2) is a factor of f(x). Perform synthetic division by 2 and expect Zero Remainder
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Bruce Mayer, PE Chabot College Mathematics
Example Example Factor Theorem Factor Theorem
SyntheticDivision
The remainder is indeed zero suggesting that (x−2) divides “evenly” into the original function; i.e. with the Quotient from the synthetic Division;
2 3 2 19 6
6 16 6
3 8 3 0
f x 3x2 2x2 19x 6 x 2 3x2 8x 3
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Bruce Mayer, PE Chabot College Mathematics
Example Example Factor Theorem Factor Theorem
ReWrite Eqn using (x−2) Factor
Use Factoring and Zero-Products to find the solution associated with the TriNomial
03832 2 xxx
3x2 8x 3 0
3x 1 x 3 0
3x 1 0 or x 3 0
x 1
3 or x 3
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Bruce Mayer, PE Chabot College Mathematics
Example Example Factor Theorem Factor Theorem
Thus the Solution for
In Set-Builder form
3x2 2x2 19x 6 0.
3,1
3, 2
.
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Bruce Mayer, PE Chabot College Mathematics
WhiteBoard WorkWhiteBoard Work
Problems From §6.5 Exercise Set• 18, 26, 42
A different Typeof SyntheticDivision
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Bruce Mayer, PE Chabot College Mathematics
All Done for TodayAll Done for Today
AnotherRemainderTheorem
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Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
AppendiAppendixx
–
srsrsr 22
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Bruce Mayer, PE Chabot College Mathematics
Graph Graph yy = | = |xx||
Make T-tablex y = |x |
-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6
x
y
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
file =XY_Plot_0211.xls
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Bruce Mayer, PE Chabot College Mathematics
x
y
-3
-2
-1
0
1
2
3
4
5
-3 -2 -1 0 1 2 3 4 5
M55_§JBerland_Graphs_0806.xls -10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
file =XY_Plot_0211.xls
xy