more about polynomials

10

10.1 Polynomials and Their Operations

10.2 Division of Polynomials

10.3 Remainder Theorem

Chapter Summary

Case Study

More about Polynomials

10.4 Factor Theorem and Its Applications

10.5 The G.C.D. and the L.C.M. of Polynomials

10.6 Algebraic Fractions

P. 2

Since we know the total volume of the pillar, we can then set up an equation containing r, where r is the radius of the sphere.However, equations involving volumes in terms of radii are usually of degree 3, which are not easy to solve.

Case StudyCase Study

We are going to learn a method to solve such kind of equations.

We need to use a polynomial to represent the volume of the pillar.

If the volume of each pillar is22 500 cm2, how can we find the radius of the spherical part?

By solving the equation obtained, we can find the radius.

P. 3

The general form of a polynomial in one variable of degree n is as follows:

10.1 10.1 Polynomials and Their Polynomials and Their OperationsOperations

A. A. Polynomials in One VariablePolynomials in One Variable

The degree of the polynomial is the same as the degree of the highest degree term. We usually write the polynomials in descending order of the degrees of terms.

,... 012

21

1 axaxaxaxa nn

nn

where (i) n is a non-negative integer,(ii) the coefficients an, an – 1, ... , a0 are

real numbers and an 0,(iii) a0 is called the constant term of the

polynomial.

In junior forms, we discovered that an algebraic expression, like x3 2x2 3x 5, is an example of polynomials in one variable.

P. 4


BB. . Addition, Subtraction and Multiplication ofAddition, Subtraction and Multiplication of PolynomialsPolynomials

Let us revise the fundamental operations of polynomials including addition, subtraction and multiplication in this section.

P. 5

Example 10.1T

Solution:7x3 11x2 6x 1

) 5x4 4x3 4x2 12



)12445()16117( 23423 xxxxxx

1244516117 23423 xxxxxx1216)411()47(5 234 xxxx

136735 234 xxxx

Add 7x3 11x2 6x 1 and 5x4 4x3 4x2 12.

5x4 3x3 7x2 6x 13

Alternative Solution:

Group the like terms in the same column and leave a space for the missing term.

P. 6

Example 10.2T



Subtract A from B means B – A.

Subtract 3x4 10x3 14x 2 from x4 6x2 15x 8.

Solution:x4 6x2 15x 8

) 3x4 10x3 14x 2 2x4 10x3 6x2 x 6

Alternative Solution:)214103()8156( 3424 xxxxxx

2141038156 3424 xxxxxx28)1415(610)31( 234 xxxx

66102 234 xxxx

P. 7

Example 10.3T



)27()47( 34 xxx )2)(47()7)(47( 3434 xxxxx

28849)147(2 345 xxxx2884972 345 xxxx

Multiply x4 7x3 4 by 7 2x.

Solution:

Alternative Solution:

x4 7x3 4

) 2x 72x5 14x4 8x

) 7x4 49x3 28 2x5 7x4 49x3 8x 28

xxxxx 814228497 4534

P. 8

For example:

The result 2x – 3 is called the quotient of the division.

10.10.22 Division of Division of PolynomialsPolynomials

AA. . Division of Polynomials by MonomialsDivision of Polynomials by Monomials

xxx 2)64( 2

x

xx

2

)32(2

32 x

In junior forms, we learnt that when a polynomial is divided by a monomial, the result can be found by cancelling out the common factor.

x

xx

2

64 2

P. 9

For example, when 12x2 17x 8 is divided by 4x 7, the above method will not work.

In this case, we have to use the method of long division.

Let us first consider the division of numbers.


BB. . Long Long Division of PolynomialsDivision of Polynomials

Note that the remainder is always less than the divisor.

Quotient DividendDivisor

Remainder

75

126577

9198713

In some cases, the method of cancelling out the common factor cannot be used in calculating the division of polynomials.

P. 10

For the division of polynomials, we can perform the long division in a similar way.

For example:To find the quotient and the remainder when 12x2 17x 8 is divided by 4x 7:



Hence we obtain the quotient 3x 1 and the remainder 1.

The first term 3x in the quotient is obtained by dividing 12x2 (the leading term of the dividend) by 4x (in the divisor) 4x 7 12x2 17x 8

3x

12x2 21x4x 8

1

4x 7 1

12x2 21x is obtained by multiplying 4x 7 by 3x.

The second term 1 in the quotient is obtained by dividing 4x (the next leading term after the first subtraction) by 4x (in the divisor)

We can stop here because the degree of the constant 1 is less than that of the divisor.

17x (21x)

P. 11

Example 10.4T

Solution:

12Quotient 2 xx

11Remainder

The missing term, 0x2, should be added into the dividend to avoid making mistakes in the calculation.



Find the quotient and the remainder when 8x3 6x 9 is divided by 4x – 2.

4x 2 8x3 0x2 6x 92x2

8x3 4x2

4x2 6x 9

x

4x2 2x 4x 9 4x 2

11

1

0x2 (4x2)

P. 12

Example 10.5T

Solution:



A divides B means B A.

Find the quotient and the remainder when 1 x 3x2 divides 4 22x2 12x3.

64Quotient x

1010Remainder x

3x2 x 1 12x3 22x2 0x 412x3 4x2 4x

18x2 4x 4

4x

18x2 6x 610x 10

6

Remainder of degree 1

Divisor of degree 2

P. 13

In the previous section, we discussed the division of two integers (987 13).


CC. . Division Algorithm of PolynomialsDivision Algorithm of Polynomials

12 75 13 987

dividend divisor quotient remainder

Actually, we have similar relationship in the division of polynomials.

The dividend can always be expressed as the sum of the product of its divisor and quotient and its remainder:

The above relationship is called the division algorithm of polynomials.

Dividend Divisor Quotient Remainder

P. 14

Example 10.6T

Solution:


CC. . Division Algorithm of PolynomialsDivision Algorithm of Polynomials

)44(divisor )1472(2784 223 xxxxxdivisor)1472(42784 223 xxxxx

)1472()42784(divisor 223 xxxxx

By long division,

The required polynomial is 2x 3.

If 4x3 8x2 7x 2 is divided by a polynomial, the quotient and the remainder are 2x2 7x 14 and –44 respectively. Find the polynomial.

2x2 7x 14 4x3 8x2 7x 424x3 14x2 28x

6x2 21x 42

2x

6x2 21x 42

3

P. 15

The notations of function we learnt in Book 4, Chapter 3 like f(x), g(x), P(x) and Q(x) can also be used to denote a polynomial.

10.10.33 Remainder TheoremRemainder Theorem

For example, P(x) x3 2x2 3x 1.

When x a, the value of the polynomial P(x) is denoted by P(a).

where 3 is the value of the polynomial when x 1.

For example,

P(1) 13 2(1)2 3(1) 1 1 2 3 1 3

P. 16

In fact, (*) is an identity which means that the equality holds for all values of x.

Substituting x a into (*), we have

When P(x) is divided by x a, we have

P(x) (x a) · Q(x) R ... (*)

where Q(x) is the quotient and R is the remainder. Since the divisor (x – a) is a linear polynomial of degree 1, the degree of the remainder must be zero, i.e., R must be a constant.

As a result, we have the remainder theorem:

Remainder Theorem When a polynomial P(x) is divided by x – a, the remainder R is equal to P(a).


Consider a polynomial P(x).

P(a) (a a) · Q(a) R 0 · Q(a) R R

By using the remainder theorem, we are able to find a remainder without actually carrying out the long division of polynomials. However, we cannot find the quotient throughout the process.

P. 17

where Q(x) is the quotient and R is the remainder.

So, by substituting , we have

Therefore, the remainder theorem can be generalized as follows:

R'xQ'nmxxP )()()(

m

nx

R'm

nQ'n

m

nm

R'm

nQ'

0

Remainder Theorem When a polynomial P(x) is divided by mx – n, the remainder R

is equal to .

m

nP


m

nP

R'

Similarly, when P(x) is divided by mx n, we have

P. 18

Example 10.7TFind the remainder when (x 2)(2x 1)(x 1) 3 is divided by (a) x 2, (b) 2x 1.


It is not necessary to expand and simplify the polynomial when applying the remainder theorem.

Let P(x) (x 2)(2x 1)(x 1) 3.

2

1P

312

11

2

122

2

1

By the remainder theorem, we have(a) remainder P(2)

(2 2)(2 2 1)(2 1) 3 0 3

3 (b) remainder

0 3

3

Solution:

P. 19

Example 10.8T


3)1( P

3)1()1()1( 23 k

311 k3k

When the polynomial x3 kx2 x is divided by x 1, the remainder is 3. Find the value of k.

Solution:Let P(x) x3 kx2 x. By the remainder theorem, we have

P. 20

Example 10.9T


3)( aP3523 2 aa0823 2 aa0)2)(43( aa

3

4or 2a

Solution:Let P(x) 3x2 2x 5.By the remainder theorem, we have

When the polynomial 3x2 2x 5 is divided by x a, the remainder is 3. Find the value(s) of a.

P. 21

Thus, if P(a) 0, that is, the remainder is 0, it means that P(x) is divisible by x a, that is, x a is a factor of P(x).

Actually, the above result is called the factor theorem of polynomials.

According to the remainder theorem, when a polynomial P(x) is divided by x a, the remainder is P(a).

Factor Theorem If P(x) is a polynomial and P(a) 0, then x a is a factor of P(x). Conversely, if x a is a factor of a polynomial P(x), then P(a) 0.

AA. . Factor TheoremFactor Theorem

10.10.44 Factor Theorem and ItsFactor Theorem and Its ApplicationsApplications

Factor Theorem

If P(x) is a polynomial and 0, then mx n is a factor of P(x).

Conversely, if mx n is a factor of a polynomial P(x), then 0.

m

nP

m

nP

The factor theorem can also be extended as follows.

P. 22

Example 10.10T

Solution:



12)1(811 23

x 1 is not a factor of P(x).

4

12)1(8)1()1( 23 12811

18

12)2(822 23 121648

Let P(x) x3 x2 8x 12.(a) Using the factor theorem, determine whether each of the following is

a factor of P(x). (i) x 1 (ii) x 1 (iii) x 2

(b) Hence factorize P(x) completely.

(a) (i) P(1)

0

(ii) P(1)

0 x 1 is not a factor of P(x).

(iii) P(2)

0 x 2 is a factor of P(x).

P. 23

Example 10.10T



128 2 23 xxxx

2x

2 23 xx 1282 xx

x

22 xx 126 x

6

126 x

)6)(2()( 2 xxxxP

)3)(2)(2( xxx

)3()2( 2 xx

Solution:(b) From (a), x 2 is a factor of P(x).

By the method of long division,

Let P(x) x3 x2 8x 12.(a) Using the factor theorem, determine whether each of the following is

a factor of P(x). (i) x 1 (ii) x 1 (iii) x 2

(b) Hence factorize P(x) completely.

P. 24

Example 10.11T

Solution:



03

1

P

023

15

3

1

3

13

23

k

023

5

99

1 k

018151 k4k

If 3x 1 is a factor of P(x) 3x3 kx2 5x 2,(a) find the value of k. (b) Hence factorize P(x) completely.

(a) Since 3x 1 is a factor of P(x), .

(b) By long division, we have P(x)

2x

254313 23 xxxx

3 23 xx 253 2 xx

x

3 2 xx 26 x

2

26 x)2)(13( 2 xxx)1)(2)(13( xxx

P. 25

We have P(x) 2x3 3x2 8x 3 (ax b)(px2 qx r),where a, b, p, q and r are integers and a 0.

The possible values of a are 1 and 2.

Consider a cubic polynomial P(x) 2x3 3x2 8x 3.Let ax b be a factor of P(x).


BB. . Applications of Applications of Factor TheoremFactor Theorem

The possible values of b are –1, –3, 1 and 3.

x – 1

x + 1

x – 3

x + 3

2x + 1 2x – 1 2x + 3 2x – 3

All possible linear factors ax + b of P(x):

Actually, we can make use of the factor theorem to verify which of the above are the factors of the polynomial P(x) 2x3 3x2 8x 3.

P. 26

Example 10.12T



Since the leading coefficient of P(x) is 1 and the constant term is 30, the possible linear factors of P(x) are x 1, x 2, x 3, x 5, x 6, etc. Do not show the unsuccessful trial in your calculation.

30)2(31)2(102 23

By the factor theorem, x 2 is a factor of P(x).

3031102 23 xxxx

2x

2 23 xx 30318 2 xx

x8

168 2 xx 3015 x

15

3015 x

)158)(2()( 2 xxxxP)5)(3)(2( xxx

Let P(x) x3 10x2 31x 30.(a) Factorize P(x). (b) Solve the equation P(x) 0.

Solution:(a) P(2)

3062408 0

By long division, we have

(b) P(x) 0(x 2)(x 3)(x 5) 0

x 2, 3 or 5

P. 27

Example 10.13T



23

18

3

15

3

13

23

23

8

9

5

9

1 285313 23 xxxx

2x

3 23 xx 286 2 xx

x2

26 2 xx 26 x

2

26 x)22)(13()( 2 xxxxP

Let P(x) 3x3 5x2 8x 2.(a) Factorize P(x) into linear or quadratic factors with integral coefficients. (b) Solve the equation P(x) 0.

Solution:

(a)

3

1P

0 By the factor theorem, 3x 1 is a factor of P(x).


P. 28

Example 10.13T



Solution:

Let P(x) 3x3 5x2 8x 2.(a) Factorize P(x) into linear or quadratic factors with integral coefficients. (b) Solve the equation P(x) 0.

(b) From (a), P(x) (3x 1)(x2 2x 2).

P(x) 0 (3x 1)(x2 2x 2) 0

3x 1 0 or x2 2x 2 0

x or 313

1

31

2

322

2

122

)1(2

)2)(1(422 2

x

P. 29

Consider a polynomial P(x) x4 3x2 2.

From the above example, we might infer that when a polynomial does not have any linear factors, or the coefficient of that factor is not rational, we cannot apply the factor theorem.


CC. . Limitation of the Use of Limitation of the Use of Factor TheoremFactor Theorem

By using the factor theorem, we can check that none of the above linear factors is a factor of P(x).

All possible linear factors of P(x) are x 1, x 1, x 2 and x 2.

In fact, P(x) be factorized as P(x) (x2)2 3(x2) 2

Therefore, if there is no linear factor in a polynomial, we cannot apply the factor theorem to factorize the polynomial.

This implies there is no linear factor in P(x).

(x2 1)(x2 2)

But this does NOT imply P(x) cannot be factorized.

P. 30

For example, consider the two numbers 30 and 42.

For two or more polynomials, we can also find their G.C.D. and L.C.M.

Greatest Common Divisor (G.C.D.) For any two or more polynomials, the common factor of them with the highest degree is called the greatest common divisor (G.C.D.).

Least Common Multiple (L.C.M.) For any two or more polynomials, the L.C.M. of them is the polynomial with the lowest degree which is divisible by each of the given polynomials.

10.5 10.5 The G.C.D. and the L.C.M. ofThe G.C.D. and the L.C.M. of PolynomialsPolynomials

In junior forms, we learnt the G.C.D. and L.C.M. of numbers.

∵ 30 2 3 5 and 42 2 3 7

∴ G.C.D. of 30 and 42 2 3 L.C.M. of 30 and 42 2 3 5 7

6 210

G.C.D. is also known as H.C.F. (highest common factor).

P. 31

Sometimes, we have to factorize the given polynomials before finding their G.C.D. or L.C.M. and factor theorem may be used if necessary.


Consider P(x) (x 2)(x 4)3 and Q(x) (x 1)(x 2)2(x 4).

G.C.D. (x 2)(x 4)

L.C.M. (x 1)(x 2)2(x 4)3

P. 32

Example 10.14T


Given two polynomials, P(x) x2 x 12 and Q(x) x3 27. (a) Factorize P(x) and Q(x). (b) Find the G.C.D. and the L.C.M. of P(x) and Q(x).

(a) P(x) x2 x 12 (x 4)(x 3)

Q(x) x3 27 (x 3)(x2 3x 9)

Solution:

(b) P(x) (x 4)(x 3) Q(x) (x 3)(x2 3x 9)

G.C.D. (x 3)

L.C.M. (x 3)(x 4)(x2 3x 9)

a3 b3 (a b)(a2 ab b2)

P. 33

Example 10.15T

613 23 23 xxxx 2 2x

62 23 xx 6135 2 xx

x5

155 2 xx 62 x

2

62 x


(a) Factorize P(x) 2x3 x2 13x 6 and Q(x) x3 2x2 9x 18. (b) Hence find the G.C.D. and the L.C.M. of P(x) and Q(x).

Solution:

By the factor theorem, x 3 is a factor of P(x).

)252)(3()( 2 xxxxP

(a) P(3) 2(3)3 32 13(3) 6


54 9 39 6 0

)2)(12)(3( xxx

Q(x) x3 2x2 9x 18 x2(x 2) 9(x 2) (x 2)(x2 9) (x 2)(x 3)(x 3)

(b) G.C.D. (x 2)(x 3)

L.C.M. (x 2)(x 3)(2x 1)(x 3)

P. 34

Example 10.16T


Find the G.C.D. and the L.C.M. of a2 ab 2b2 and a2 4b2.

∵ a2 ab 2b2 (a b)(a 2b)

∴ G.C.D. (a 2b)

Solution:

a2 4b2 (a 2b)(a 2b)

L.C.M. (a b)(a 2b)(a 2b)

P. 35

10.10.6 Algebraic Fractions6 Algebraic Fractions

)(

)()(

xQ

xPxR

2

1

x 3

42

xx

xAn algebraic fraction is a quotient of two polynomials where the

denominator is not equal to zero, such as and .

Remark: If a function R(x) is defined as the quotient of two non-zero polynomials

P(x) and Q(x), i.e., , we call the function a rational function.

P. 36

)2)(1(

)3(4

)3(9

)1)(2(

xx

x

x

xx


A. Multiplication and Division of AlgebraicA. Multiplication and Division of Algebraic FractionsFractions

The multiplication and division of algebraic fractions are just like the multiplication and division of fractions that we can just cancel the common factors.

For example:

)1(

4

9

)1(

x

x

)1(9

)1(4

x

x(x 2) and (x 3) are the common factors of the numerator and denominator.

P. 37

Example 10.17T

Solution:

223

1 2

2

2

x

xx

xx

x

2

)1(

)1)(2(

)1)(1(

x

xx

xx

xx

)1(

2

)1)(2(

)1)(1(

xx

x

xx

xx

x

1



Simplify .223

1 2

2

2

x

xx

xx

x

First factorize the expressions. Then cancel the common factors.

P. 38

Example 10.18T

)2)(( baba



Solution:

Simplify .22

22 24

ba

ab

ba

ba

22

22 24

ba

ab

ba

ba

))((

2)2)(2(

baba

ba

ba

baba

ba

baba

ba

baba

2

))(()2)(2(

P. 39


BB. . Addition and Subtraction of AlgebraicAddition and Subtraction of Algebraic FractionsFractions

Algebraic fractions can be added or subtracted.

When the denominators of two algebraic fractions are equal, we can add or subtract them.

However, when the denominators are not the same, we have to make the denominators equal which can be done by finding the L.C.M. of the denominators.

P. 40

Example 10.19T

2)2(

)2(3

x

xx

2)2(

62

x

x

2)2(

)3(2

x

x

2)2(2

3

x

x

x



Solution:

Simplify .442

32

xx

x

x

442

32

xx

x

x

22 )2()2(

)2(3

x

x

x

x

P. 41

Example 10.20T

)1)(3(

1

)3)(3(

3

xx

x

xx

)3)(3)(1(

)3)(1()1(3

xxx

xxx

)3)(3)(1(

62

xxx

xx

)3)(3)(1(

)2)(3(

xxx

xx

)3)(1(

)2(

xx

x



Solution:

Simplify .32

1

9

322

xx

x

x

32

1

9

322

xx

x

x

Expand and simplify.

The L.C.M. of (x 3)(x 3) and (x 3)(x 1) is(x 1)(x 3)(x 3).

Factorize the numerator.

P. 42

10.1 Polynomials and Their Operations

A polynomial in one variable of degree n can be expressed as

where an, an – 1, ... , a0 are real numbers and an 0.

Chapter Chapter SummarySummary

,... 012

21

1 axaxaxaxa nn

nn

P. 43

1. If a polynomial P(x) is divided by a polynomial Q(x), we can

use the method of long division to find the quotient and the remainder.


2. When a polynomial P(x) is divided by a divisor, we get a quotient and a remainder.

The division algorithm of polynomials is written as

dividend divisor quotient remainder.

10.2 Division of Polynomials

P. 44

1. When a polynomial P(x) is divided by x a, the remainder R is equal to P(a).


m

nP

2. When a polynomial P(x) is divided by mx n, the remainder R is

equal to .

10.3 Remainder Theorem

P. 45

1. If P(x) is a polynomial and P(a) 0, then x a is a factor of P(x). Conversely, if x – a is a factor of P(x), then P(a) 0.


2. If P(x) is a polynomial and , then mx n is a factor of P

(x). Conversely, if mx n is a factor of P(x), then .

0

m

nP

0

m

nP

10.4 Factor Theorem and Its Applications

P. 46

1. For any two or more polynomials, the common factor of them with the highest degree is called the greatest common divisor (G.C.D.).


2. For any two or more polynomials, the common multiple of them with the least degree is called the least common multiple (L.C.M.).

10.5 The G.C.D. and the L.C.M. of Polynomials

P. 47

An algebraic fraction is a quotient of two polynomials where the denominator is not equal to zero. We can perform addition, subtraction, multiplication and division for algebraic fractions.

Chapter Chapter SummarySummary10.6 Algebraic Fractions

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