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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

[email protected]

Chabot Mathematics

§6.5 Synthetic§6.5 SyntheticDivisionDivision



Review §Review §

Any QUESTIONS About• §6.4 → PolyNomial Long Division

Any QUESTIONS About HomeWork• §6.4 → HW-26

6.4 MTH 55



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.



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



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.




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.




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



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




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.



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



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.



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



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.




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



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




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.




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.



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.



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.



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



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



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



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



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




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




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




Thus the Solution for

In Set-Builder form

3x2 2x2 19x 6 0.

3,1

3, 2

.



WhiteBoard WorkWhiteBoard Work

Problems From §6.5 Exercise Set• 18, 26, 42

A different Typeof SyntheticDivision



All Done for TodayAll Done for Today

AnotherRemainderTheorem



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



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



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

[email protected] mth55_lec-33_sec_6-5_synthetic_division.ppt 1 bruce mayer, pe chabot...

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