[email protected] MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 1 Bruce Mayer, PE Chabot College...

[email protected] • MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§6.4 Divide§6.4 DividePolyNomialsPolyNomials



Review §Review §

Any QUESTIONS About• §6.3 → Complex Rational

Expressions

Any QUESTIONS About HomeWork• §6.3 → HW-25

6.3 MTH 55



§6.4 Polynomial Division §6.4 Polynomial Division

Dividing by a Monomial

Dividing by a BiNomial

Long Division



Dividing by a MonomialDividing by a Monomial

To divide a polynomial by a monomial, divide each term by the monomial.

EXAMPLE – Divide: x5 + 24x4 − 12x3 by 6x

Solution5 4 3 5 4 324 12 24 12

6 6 6 6

x x x x x x

x x x x

5 1 4 1 3 11 24 12

6 6 6x x x

4 3 214 2

6x x x



Example Example Monomial Division Monomial Division

Divide: 5 4 3 2 2 221 14 7 7a b a b a b a b

Solution: 5 4 3 2 2 5 4 3 2 2

2 2 2 2

21 14 7 21 14 7

7 7 7 7

a b a b a b a b a b a b

a b a b a b a b

5 2 4 1 13 2221 14 7

7 7 7a b a b

123 33 abba



Dividing by a BinomialDividing 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.



Recall Arithmetic Long DivisionRecall Arithmetic Long Division

Recall Whole-No. Long Division

Divide: 157

12 157

12

37

36

1

13

1

2

Divisor Quotient

Remainder

Quotient 13

Divisor 12

Remainder 1

= Dividend= 157

•++

•



Binomial Div. Binomial Div. Step by Step Step by Step

Divide 2x³ + 3x² - x + 1 by x + 2

3 22 2 3 1x x x x x + 2 is the divisor

The quotient will be here.

2x³ + 3x² - x + 1 is the dividend

Use an IDENTICAL Long Division process when dividing by BiNomials or Larger PolyNomials; e.g.;




First divide the first term of the dividend, 2x³, by x (the first term of the divisor).

3 22 2 3 1x x x x

22xThis gives 2x². This will be the first term of the quotient.




Now multiply (x+2) by 2x²

3 22 2 3 1x x x x 3 22 4x x

22x

2xand subtract




Bring down the next term, -x.

3 22 2 3 1x x x x 3 22 4x x

22x

2x x




Now divide –x², the first term of –x² - x, by x, the first term of the divisor

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

which gives –x.




Multiply (x +2) by -x

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

xand subtract




Bring down the next term, 1 x

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

1




Divide x, the first term of x + 1, by x, the first term of the divisor

13 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

x 1which gives 1




Multiply x + 2 by 1

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

x

1

12x 1and subtract




The remainder is –1.

3 22 2 3 1x x x x 3 22 4x x

22x

2x x

x

2 2x x

x

1

12x 1

The quotient is 2x² - x + 1



Example Example BiNomial Division BiNomial Division

Divide x2 + 7x + 12 by x + 3.

Solution

2

2

3 7 12

( )

3

4

x

x

x x x

x

x

Subtract by changing signs and adding

Multiply (x + 3) by x, using the distributive law




Solution – Cont.

2

2

4

+ 3 7 12

( 3 )

12

( )

2

0

4

1

4

x

x

xx x x

x x

Subtract

Multiply 4 by the divisor, x + 3, using the distributive law

Bring Down the +12




Divide 15x2 − 22x + 14 by (3x − 2) Solution

2

2

3 2 15 22 14

( )

12 14

( 12 8)

6

15 1

5 4

0

x

x

x

x

x

x

x x

The answer is 5x − 4 with R6. We can also write the answer as:

65 4 .

3 2x

x




Divide x5 − 3x4 − 4x2 + 10x by (x − 3) Solution

The Result

5 4 3 2

5 4

3 2

2

3 3 0 4 10 0

( 3 )

0 4 10

4 12

2 0

( 2 6)

x x x x x x

x x

x x x

x x

x

x

4 4 2

6

x x

4 6

4 2 .3

x xx



Divide 3Divide 3xx22 −− 4 4xx −− 15 by 15 by xx −− 3 3

SOLUTION: Place the TriNomial under the Long Division Sign and start the Reduction Process

23 3 4 15x x x 3x

2(3 )9x x

5 15x

Divide 3x2 by x: 3x2/x = 3x.

Multiply x – 3 by 3x.

Subtract by mentally changing signs and adding −4x + 9x = 5x.



Divide 3Divide 3xx22 −− 4 4xx −− 15 by 15 by xx −− 3 3 SOLUTION: next divide the leading

term of this remainder, 5x, by the leading term of the divisor, x.

23 3 4 15x x x 3 5x

2(3 9 )x x

5 15x

Divide 5x by x: 5x/x = 5.

Multiply x – 3 by 5.

Subtract. Our remainder is now 0.

5(5 1 )x

0

CHECK: (x − 3)(3x + 5) = 3x2 − 4x − 15 The quotient is 3x + 5.



Formal Division AlgorithmFormal 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 thatF(x) = D(x) • Q(x) + R(x)

Dividend Divisor Quotient Remainder

Either R(x) is the zero polynomial, or the degree of R(x) is LESS than the degree of D(x).



PolyNomial Long DivisionPolyNomial 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.



PolyNomial Long DivisionPolyNomial 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.



Example Example TriNomial Division TriNomial Division

Divide x4 13x2 x 35 by x2 x 6.

SOLN

x2 x 6 x4 0x3 13x2 x 35

x4 x3 6x2

x3 7x2 x 35

x3 x2 6x

6x2 7x 35

x2 x




SOLNcont. x2 x 6 x4 0x3 13x2 x 35

x4 x3 6x2

x3 7x2 x 35

x3 x2 6x

6x2 7x 35

6x2 6x 36

x 1

x2 x 6




Divide x4 13x2 x 35 by x2 x 6.

The Quotient = x2 x 6.

The Remainder = x 1.

Write the Result in Concise form:

x4 13x2 x 35

x2 x 6x2 x 6

x 1

x2 x 6.



WhiteBoard WorkWhiteBoard Work

Problems From §6.4 Exercise Set• 30, 32, 40

BiNomialDivision



All Done for TodayAll Done for Today

PolynomialDivisionin base2

From UC Berkeley Electrical-Engineering 122 Course



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

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