Senior 3 Pre-Calculus Mathematics Module 8, Lesson 3 25

Lesson 3

Remainder Theorem and Factor Theorem

Objectives

1. You will learn how to divide polynomials using long divisionand synthetic division.

2. You will learn how to apply the remainder and factortheorems.

You have had practice in adding, subtracting, and multiplyingpolynomials. Dividing polynomials has many importantapplications and is especially valuable in factoring and findingthe zeros of polynonual functions

There are two algorithms for polynomial division: long divisionand synthetic division. We will examine long division first andthe relate it to synthetic division

Example 1

Dividex2—6--xbyx ÷ 2.

Solution

x —3 quotient

divisor x + 2)x2— x —6 dividend

V x2+2x—3x—6

V

—3x—600

The answer to the division is the quotient.

1. Arrange the dividend and the divisor in descending powers ofthe variable. Notice the powers of x go from 2— 1 —0.

2. Insert with 0 coefficients any missing terms.

3. Divide the first term of the divisor into the first term of the(2 ‘

dividend I = x I and place the answer over the second term.Lx )

26 ModuleS, Lesson 3 SenIor 3 Pre-CalculuS Mathematics

4. Multiply the divisor by x, line up the terms, subtract, and bringdown the —6.

5. Repeat steps 3 and 4 until the degree of the remainder is lessthan that of the divisor.

Note: In this example there is no remainder.

We write the result as:

x2—x—6=x+2

x+2

Example 2V

Findthequotientof4x3—3x+5 ÷x—3.V

4x2 + 12x * quotientdivisor x —3)4x3 + Ox2 — 3x +5 4 dividend

V4x3-12x2 [12x2 —3x

V 12x2—36x EV 33x+5V

V

33x —109remainder 114

1. Rearrange the dividend in descending powers andVfor the missing term in the sequence.

2. Divide x into 4x3 = 4x2] and multiply 4x2 by the

divisor, line up terms, subtract, and bring down —3x.3 Repeat step 2

V

dividendquotientV

A 114 - remamder‘iX X +LP

= 4x2 +12x +33 +

_____

x—3 x—3 divisorV V

divisor

or4x—3x÷5=(x—3)(4x2+12x+33)+ 114dividend = divisor x quotient + remainder

0

Senior 3 Pre-Calculus Mathematics Module 8, Lesson 3 27

Example 33 2Divide6x +16x—19x —4 ÷x—2.

Solution

6x2 — 7x +2x _2)6x3 —19x2 +16x —4

6x3—12x2

—7x2+16x

—7x2+14x2x—4

2x—4

00

6x3+16x—19x2—4=(x—2)(6x3+7x+2)+Odividend = divisor x quotient + remainder

The above examples illustrate the Division Algorithm.It states:

If/tx) and d(x) are polynomials such that d(x) 0, and the degreeof d(x) is less than or equal to the degree offix), then there arepolynomials q(x) + r(x) such thatf(x) = d(x)q(x) +r(x) where r(x) = 0dividend = (divisor)(quotient) ÷ remainder or the degree

of r(x) is lessorthan thedegree of d(x).

f(x) =(x—a)q(x)represent the same relationship

If the remainder r(x) =0, then d(x) divides evenly into fix).When the divisor is of the form x ± a, synthetic division can beused as a shortcut for long division using polynomials.

28 Module B, Lesson 3 SenIor 3 Pre-Calculus Mathematics

When doing synthetic division, the terms of the dividend must

be arranged in descending order of powers and any missing

terms replaced with a zero. You will use only the coefficients of

the terms to do the division. The divisor has to be of the form

x ± a. Ifx — a is the divisor, a is positive. If x + a is the divisor,

treat it as x — (—a) and use a as a negative value.

Example 4

divisor — 2)6 —19 . 16 —4 dividendcoefficients

12 —14 v 4

6 2x6 —7 2x(—7) 2 2x2 0

coefficients of the remainder

quotients

Quotient: 6x2 — 7x + 20 and the remainder is zero.

Solution

Bring down the six, multiply (2) x 6, place 23 below the —19.

Add —19 + 12 = —7, multiply 2 x (—7), and place —14 below 16.

Add 16+ (—14) =2.

Multiply 2 x 2 and place the 4 under the —4.

Add.

...6x2—19x+16x—4=(x—2)(6x2—7x+20)

The key steps of the division are:

1. Arrange the coefficients of f(x) in order of descending powers

of x (write 0 as the coefficient for each missing power).

2. After writing the divisor in the form x — a, use a to generate

the second and third rows of numbers as follows. Bring down

the first coefficient of the dividend and multiply it by a: then

add the product to the second coefficient of the dividend.

Multiply this sum by a, and add the product to the third

coefficient of the dividend. Repeat the process until a product

is added to the constant term of/(x).

3. The last number in the third row of numbers is the

remainder; the other numbers in the third row are the

coefficients of the quotient, which is of degree 1 less than fix).

I

Senior 3 Pre-Calculus Mathematics Module 8 Lesson 329

Example 5Use synthetic division to divide x4 — lOx2 — 2x + 4 by x + 3.SolutionDivisor: x + 3 = x — (—3).divisor —* —3)1 0 —10 —2 3 —* dividend coefficents

—3 —1 3 —3 and missing term1 —3 —1 1 1 coefficients

quotient coefficients remainder

x4—10x2—2x+4=x3—3x2—lx+1+ 1

or

x4 —lOx2—2x+4 =(x÷3)(x —3x2 —Lx

The remainder obtained in the synthetic division process has aninteresting connection to polynomial functions.Exampie 6

4 2Evaluate fix) = x — lOx — 2x ÷4 at fi—3).Solution

f(-3) = (_34) 10(3)2 2(-3) +4= 81—90+6÷4=1

Notice fi—3) = 1 which represents the remainder fromExample 5.

If a polynomial fix)) is divided by x — a, then the remainder isr =

From the Division Algorithm, we have fix) = (x — a) q(x) + Rwhere R is the remainder.

If we fInd fia)=(a —a)g(x)+ Rfia) = R which represents the remainderWhen you are interested in the value of the remainder, you willuse the Remainder Theorem to obtain the value of theremainder.

‘1]

30 Module 8, Lesson 3 SenIor 3 Pm-Calculus Mathematics

Example 7 0Find the remainder when x4 — 2x3 + 5x + 2 ÷ x + 1.

Solution

By the Remainder Theorem, 1(a) = R.

x + 1 is rewritten as x — (—1) to determine that a = —1.

f(x)=x4—2x3+5x+2

f(-i) = (-i) - 2(-1) + 5(-1) +2

1+ 2-5+ 2

f(1)=O

When the remainder is 0, what relation must exist between the

divisor and the dividend?

It means that the polynomial is evenly divisible by the divisor.

Taken one step further, it means that x + 1 must be a factor of

1(x). This can be developed by exanining the Factor Theorem.

it states: LA polynomial fix) has a factor (x — a) if and only ifjta) =0.

Using the Division Algorithm with the factor x — a, we have

f(x)=(x—a)q(x)+f(a).

By the imai der Theorem, 1(a) = R so - L/(x)=(x—a)q(x)÷f(a).

Since 1(a) =0then/(x)=(x—a)q(x)

which shows that x — a is a factor of1(x).

Conversely, ifx — a is a factor offix), then dividing fix) by x — a

yields a remainder of 0.

T

SenIor 3 Pre-Calculus MathematIcs Module 8, Lesson 3 31

Example 8

a) Determine whether x + 2 is a factor offix) = x3 — 6x —4.

b) Determine the other factors of/tx).

Solution

a) x + 2 will be a factor iffi—2) = 0

f(x) =x3—6x—4

f(—2) = (—2) — 6(—2) —4

= —8+12—4

=0

Since its remainder is 0, x + 2 must be a factor.

b) Do synthetic division to find the quotient that will be a factor.

—2)1 0 —6 —4

—2 4 4

1 —2 —2 0

The quotient is x2 — 2x —2.

fix) = (x + 2)(x2— 2x — 2). These are the factors.

Ifx2 — 2x —2 had been factorable, then you would have listed

its factors along with x + 2 in your answer.

Example 9

Factor completely: fix) = x3 — 2x2 — 5x + 6.

Solution

When you examine this question, it is not apparent what choices

you should make to determine a factor of the form x — a.

The choices can be listed as rational numbers whose numerators

are factors of the constant term (term without a variable in it)

and whose denominators are factors of the leading coefficient

which is the coefficient of the term having the highest degree.

factors of constant term

factors of leading coefficemt

32Module 8 Lesson 3 SenIor 3 Pre-Calculus Mathematics

•JEC

Once you have listed the possible “a” values, you can use theRemainder Theorem and Factor Theorem and/or syntheticdivision to obtain the factors.Note: When the leading coefficient is 1, then the possible “a”values are simply the factor of the constant term.

3 2fix)=x —2x —5x+6

Because the coefficient ofx3 is 1, the possible “a” values are thefactors of 6.

a = ±1, ±2, ±3, ±6

Choose a = 1Iffla) = 0, then x — a is a factor

f(1)=13_2(1)2 —5(1)÷6=1—2—5+6=0

4

:

L

Fr

;.x—lisafactor

Use synthetic division to get the second factor:1)1 —2 —5 6

1 —1 —61-—i. —6 0

quotient isx2—x—6

Notice that this is a quadratic that will factorx2—x--6=(x--3)(x÷2)

x3 — 2x2 — 5x +6= (x — 1)(x — 3)(x +2)

Li

[

Senior 3 Pro-Calculus Mathematics Module 8, Lesson 3 33

Example 10

Factor: 2x3 + 3x2 — 8x + 3.

Solution

Since the leading coefficient is 2 and the constant term is 3, the

possible a values are:

a= factors of 3 = ±1, ± 3

= ±1, ± 3, ± ±factors of 2 ±1,±2 2 2

Check if a = 1

f(1)=2.13+2.12—8(1)+3

=2+3—8+3

=0

:.x—lisafactor

Use synthetic division to get a second factor:

1)2 3 —8 3

2 5—3

2 5—3 0

Secondfactor: 2x2+5x—3

This will factor: 2x2 + 5x —3 = (2x — 1)(x + 3)

Factors are: (x — 1)(2x — 1)(x + 3)

34 ModuleS, Lesson 3 SenIor 3 Pre-Calculus Mathematics

Assignment 31. Divide using long division and write in the form given by the

Division Algorithm. [a) 2x3—3x+1÷x—2b)2x2+x3—3x—4÷x+2 Lc)6x3—16x2+17x—6÷3x--2d)2x3+3x2—7—4x÷x2—2e)x4—2x3—7x2+8x+12÷x+1

2. Use synthetic division to find the quotient and theremainder.

a)x3—7x+6÷x—2b) 3x4—x—4÷x—2c)x4—2x3—70x+20÷x—5d)2x3—5x2+6x+3÷2x—1e)4x3+4x2—7x—6÷2x+3

3. Verifr whether or not x + 2 is a factor of/tx) = x3 +4x2—8.4. Factor completely:

a) f(x)=x38x2±4x+48

b) f(x)=x4—2x3—17x2+l8xi-72

c) f(x) = —27

d) f(x)=12x3—12x2+3x—3

5. When a polynomial fix) is divided by 2x + 1, the quotient isx2 — x + 4 and the remainder is 3. Find fix).

C

SenIor 3 Pro-Calculus Mathematics Module 8, Lesson 3 35

6.

—1

The volume of the rectangle prism isV=3x3-i-8x2—45x—5O.

Find the missing dimension.

7. The polynomial p(x) = 4x3 -i-bx2 ÷ cx + 11 has a remainder of—7 when divided by Cx +2) and a remainder of 14 whendivided byx —1.

• UU

36 Module 8, Lesson 3 SenIor 3 Pie-Calculus Mathematics

Notes 0TIU00

• •Z ••• •••

C

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Senior 3 Pre-Calculus Mathematj Module 8, Lesson 3, Answer Key15

Answer Key

Lesson 31. a) 2x2 + 4x +5 Answer

x—2x+o3x+1(x_2)(2x2+4x+5)+11

2x3—4x2•

. 4x2—3x

28x

5x+.1

fl . 5x—1O

11

b) x2—3 Answerx+2÷2x2_3x

(x÷2)(x2_3)+22x2

—3x—4

rS-6

c) 2x2 — 4x +3 Answer

fj 3x — 2x — 16x .+ 17x —6(3x — 2)(2X2 4x +3)

• 6x3—4x2

U 12x2+17x

—12x2-i-8x

9x—6U9x—6

cid) 2x +3 Answer

x2— +3x2::

_7(x2

— 2)(2x + 3)—i

U 3x2 —7

3x2 —6

16 Module B, Lesson 3, Answer Key Senior 3 Pre-Calculus Mathematics

e) — 3x2 — 4x + 12 swer

x+i)x4 —2x8—7x2 +8x+12 (x+1)(x —3x2 _4x+12)

4

::::;::—4x2+8x

—4x2—4x

12x+12

12x+12 fl0

2. a) 2)1 0 —7 6 quotient: x2 + 2x —3 U2 4 —6 remainder: none

12—3 0

b) 2)3 0 0 —1 —4 quotient:3x3+6x2+12x+23

6 12 24 46 remainder: +

3 6 12 23 42

c) 5)1 —2 0 —70 20 quotient: r5 15 75 25 x3+3x2+15x÷5

451 3 15 5 45 remainder: +

x—5

d) Before you do this question, 2x — 1 has to be changed mto1

its equivalent form X —

.

F

—5 6 3quotient: 2x2 — 4x +4

L

2 . 5remainder: +

2-445 2x_1,:

Senior 3 Pre-Calculus Mathematics Module 8, Lesson 3, Answer Key 17

e)..)4 4 —7 —6 quotient: 4x2 + lOx +82

. 61 remainder: +

_______________

2x+34 10 8 6

3. You can do this in one of three ways:a) Synthetic division

—2)1 4 0 —8

—2 —4 +8

1 2 —4 0 +-remainder

Because the remainder is 0, then x + 2 is a factor.b) Long division

x2 +2x—4x+2)x3+4x2+0_8

x3-i-2x2

2x2+022x +4x

—4x—8fl —4x—8

0

Because the remainder is 0, then x ÷ 2 is a factor.If f(—2) = 0, then x + 2 is a factor

f(x)=x3÷4x2—8

f(—2) = (—2) + 4(2)2 —8

U =—8+16—8

Since the remainder is 0, then x + 2 is a factor.

U

18 Module 8. Lesson 3, Answer Key Senior 3 Pre-Calculus Mathematics

c) Use the Remainder Theorem

f(—2) gives the remainder

f(—2) (—2) + 4(_2)2 —8

=—8+16—8

=0

.. x + 2 is a factor when the remainder is 0

4. a) f(x)=x3—8x2+4x+48

factors of constant termPossible “a values =

factors of leading coefficient

= ±1,±2, ±3,±4, ±6, ±8, ±12,

±16, ± 24, ± 48

Note: Since the leading coefficient is 1, then the factors

are ±1.

To find a remainder, let f(a) = 0.

f(x)=x3—8x2+4x+48

f(1)=13_8(1)2 +4(1)÷48

=1—8+4÷48

=—45 ... x—lisnotafactor

f(2)=2—8(2)2-i-4(2)+48

=8—32+8+48

=32 .. x—2isnotafactor

f(3)=33_8(3)2 +4(3)÷48

= 27—72+12+48

=—5 :.x—3isnotafactor

f(4) =43_8(4)2+4(4) +48

= 64—128+16+ 48

=0 .. x—4isafactor

Ufl

Senior 3 Pre-Calculus Mathematics Module 8, Lesson 3, Answer Key19

Use synthetic division to find a second factor4)1 —8 4 48

U 4 —16 —48

.. x2 —4x — l2is the o:her factor which will factor furtherthe factors are (x — 4)(x2 — 4x — 12) or{J V (x —4) (x — 6)(x + 2)

b) f(x) = — 2x3 — 1 7x2 + 18x + 72fl-

,, factors of constant termPossible a values =factors of leadmg coefficientI

.V = ±1, ±2, ±3, ±6, ±8, ±9, ±12,

±18, ± 24, ± 36, ± 72

f(1)=14—2-1 _17(1)2÷1.8.1÷72V =1—2—17+18+72

=72 x --1 is not a factor

f(—1) = (—i) — 2(—1) — 17(_1)2 + 18(—1) +7217=1÷2—17—18÷72

Q =40 x +1 is not a factor

1(2) = (2) —2- 2 — 17(2)2 + 18(2)+ 72U =16—16—68+36+72

.U =0 x +2 is not a factor

f(—2) = (—2) — 2(—2) — 17(_2)2 + 18(—2) +720 =16÷16—68—36+72=0 :.x+2isafactor

o3

....

...

,•

20 Module 8, Lesson 3, Answer Key Senior 3 Pre-Calculus Mathematics

Use synthetic division to find another factor:

—2)1 —2 —17 18 72

—2 8 18 —72

1 —4 —9 36

— 4x2 — 9x + 36 can still be a factor

f(3)3—4(3)2—9.3÷36

=27—36—27÷36

=0 ..x—3isafactor

Use synthetic division again use x = 3 and

x3—4x2—9x+ 36.

3)1 —4 —9 36

3 —3 —36

1 —1 —12 0

9 27

q

0

quotient is x2 — x — 12 which factors into (x — 4)(x + 3)

thefactorsare(x+2)(x—3)(x+3)(x—4)

3c) f(x)x —27

Possible “a” factors: ±1, ±3, ±9, ±27

f(3)=33—27

=27—27= 0 x —3 is a factor

Do synthetic division:

3)100 -27

3

139 0

Other factor is x2 + 3x + 9— does not factor further.

factors are (x — 3) (x2 + 3x + 9)

F

JE

[Cr

Senior 3 Pre-Calculus Mathematics Module 8, Lesson 3, Answer Key21

d) f(x)3(4x3—4x2+x—1)

factors of constant termPossible a values =factors of leading coefficient

+1=— =+1+_+_±1,±2,±4 ‘2’4

f(1) = 3(4(1) - 4(1)2 + 11)

=3(4—4+1—1)fl = 0 :. x —1 is a factor

Use synthetic division to get another factor.U 1)4 —4 1 —1

fl. 4010

The three factors are 3(4x2 + 1)(x — 1)5. f(x)=(2x+1)(x2_x+4)+311

=2x3—2x2+8x+x2—x+4+3=2x3—x2÷7x+7U

6. The volume of the rectangular prism isV 3x3+8x2—45x—50.Use synthetic division with V and x + 1.—1)3 8 —45 —50

—3 —5 +503 5 —50 0

U.. The quotient is 3x2 + 5x —50 which factors into(3x—10)(x+5)

The missing term is 3x — 10.

U3.

..,,., .

22Module 8, Lesson 3, Answer Key Senior 3 PrealcuIUs Mathematics C

f(—2) = 4(—2) + + c(—2)+11 which equals —7

_32+4b—2X117

4b—2c=14 (1)

f(1) = 4(1) + b(1)2 + c(1) + 11 which equals 14 D144+b+c+11

b+c+1514fl

b+c—1 (2)

4b — 2c =14

b+c=—l

4b—2c=14

2b+2c—2 Eqn(2)X2

6b=12U

c=—3

p(x)=4x3+2X23X+11

t

tr