25
Module 8, Lesson 3
Senior 3 Pre-Calculus Mathematics
Lesson 3
Remainder Theorem and Factor Theorem
Objectives
1. You will learn how to divide polynomials using long division
and synthetic division.
2. You will learn how to apply the remainder and factor
theorems.
You have had practice in adding, subtracting, and multiplying
polynomials. Dividing polynomials has many important
applications and is especially valuable in factoring and finding
the zeros of polynonual functions
There are two algorithms for polynomial division: long division
and synthetic division. We will examine long division first and
the relate it to synthetic division
Example 1
—6--xbyx ÷ 2.
2
Dividex
Solution
x + 2)x2
divisor
—
2x
x
+
2
—3x—6
—3x—6
00
V
quotient
x —3
x —6
dividend
V
The answer to the division is the quotient.
1. Arrange the dividend and the divisor in descending powers of
the 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
Lx
‘
=
x I and place the answer over the second term.
)
26
ModuleS, Lesson 3
SenIor 3 Pre-CalculuS Mathematics
4. Multiply the divisor by x, line up the terms, subtract, and bring
down the —6.
5. Repeat steps 3 and 4 until the degree of the remainder is less
than that of the divisor.
Note: In this example there is no remainder.
We write the result as:
x
—
2
x—6
=x+2
x+2
Example 2
Findthequ
—
3
3x
otie+
ntof4x
5 ÷x—3.
*
2 + 12x
4x
divisor
4
x —3) 4x
3 + Ox
2 3x +5
V
—
V
quotient
dividend
2
4
3
12
x
x
2 —3x
12x
12
—
2
36x
x
33x+5
33x —109
114
V
V
V
remainder
[
E
V
V
1. Rearrange the dividend in descending powers and
for the missing term in the sequence.
2. Divide x into 4x
3
= 4x2]
V
and multiply 4x
2 by the
divisor, line up terms, subtract, and bring down —3x.
3 Repeat step 2
V
dividend
quotient
V
A
‘iX
V
V
X
x—3
+LP
=
2 +12x +33
4x
+
114
x—3
-
remamder
divisor
divisor
or
4x—3x÷5=(x—3)(4x 12x+33)+ 114
+
2
dividend = divisor x quotient
+ remainder
0
Senior 3 Pre-Calculus Mathematics
Module 8, Lesson 3
Example 3
Divide6x 3 +16x—19x 2 —4 ÷x—2.
Solution
2 7x +2
6x
x _2)6x3 2
—19x +16x —4
—
2
—
3
6x
12x
2
—7
+
16x
x
—7
+
2
14x
x
2x—4
2x—4
00
6
+
3
—
2
+
16
7x
4=(x—
x—
+x
2)
2)(
19
+O
6x
x
dividend
= 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 degree
of d(x) is less than or equal to the degree of fix), then there are
polynomials q(x) + r(x) such that
f(x)
= d(x)q(x)
+r(x)
where r(x) = 0
or the degree
dividend
= (divisor)(quotient) ÷ remainder
of
r(x) is less
or
than the
degree 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 be
used as a shortcut for long division using polynomials.
27
SenIor 3 Pre-Calculus Mathematics
Module B, Lesson 3
28
must
When doing synthetic division, the terms of the dividend
ing
be arranged in descending order of powers and any miss
ients of
terms replaced with a zero. You will use only the coeffic
the terms to do the division. The divisor has to be of the form
or,
x ± a. If x a is the divisor, a is positive. If x + a is the divis
treat it as x (—a) and use a as a negative value.
—
—
Example 4
divisor
—
—19
2)6
.
12
6 2x6
—7 2x(—7)
16
—4
—14
v4
2 2x2
dividend
coefficients
0
remainder
coefficients of the
quotients
2 7x + 20 and the remainder is zero.
Quotient: 6x
Solution
Bring down the six, multiply (2) x 6, place 23 below the —19.
16.
Add —19 + 12 = —7, multiply 2 x (—7), and place —14 below
Add 16+ (—14) =2.
Multiply 2 x 2 and place the 4 under the —4.
—
Add.
x
)
2)(6
20
4=(x—
+
xx
7x—
19x+16
6
—
—
2
...
The key steps of the division are:
powers
1. Arrange the coefficients of f(x) in order of descending
of x (write 0 as the coefficient for each missing power).
generate
2. After writing the divisor in the form x a, use a to
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
uct
coefficient of the dividend. Repeat the process until a prod
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 3
29
Example 5
Use synthetic division to divide x
4 — lOx
2 — 2x
+
4 by x
+
3.
Solution
Divisor: x
+
3
=
x — (—3).
divisor —* —3)1
1
0
—10
—2
3
—3
—1
3
—3
—3
—1
1
quotient coefficients
—*
dividend coefficents
and missing term
1
coefficients
remainder
—
4
x
—
2
—
3
—
2
lx
x
1
+
3
+
0
4
1
x
=
x
+
x
1
or
4 —l
x
2
Ox —2x+4 =(x÷3)(x —3
2
x —Lx
The remainder obtained in the synthetic
division process has an
interesting connection to polynomial fun
ctions.
Exam pie 6
Evaluate fix) = x 4 — lOx 2 2x ÷4 at fi—3
).
—
Solution
f(-3) = (_34)
10(3)2
2(-3) +4
81—90+6÷4
=1
=
Notice fi—3)
Example 5.
=
1 which represents the remainder from
If a polynomial fix)) is divided by x —
a, then the remainder is
r=
From the Division Algorithm, we have
fix)
where R is the remainder.
=
(x — a) q(x)
+
R
If we fInd fia)=(a —a)g(x)+ R
fia) = R which represents the remainder
When you are interested in the value of
the remainder, you will
use the Remainder Theorem to obtain
the value of the
remainder.
‘1]
SenIor 3 Pm-Calculus Mathematics
Module 8, Lesson 3
30
Example 7
4
Find the remainder when x
0
—
3
2x
+
5x + 2 ÷ x
+
Solution
By the Remainder Theorem, 1(a) = R.
x + 1 is rewritten as x (—1) to determine that a
—
1.
=
—1.
2
x
x
+
=
2
x)
5x
3
f(
—
4
+
f(-i) = (-i) 2(-1)
1+ 2-5+ 2
f(1)=O
-
+ 5(-1) +2
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:
A 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).
L
—
—
By the imai der Theorem, 1(a)
/(x)=(x—a)q(x)÷f(a).
=
R so
L
-
Since 1(a) =0
then/(x)=(x—a)q(x)
which shows that x
—
a is a factor of 1(x).
Conversely, if x a is a factor offix), then dividing fix) by x
yields a remainder of 0.
—
—
a
T
31
Module 8, Lesson 3
SenIor 3 Pre-Calculus MathematIcs
Example 8
a) Determine whether x + 2 is a factor of fix)
b) Determine the other factors of/tx).
Solution
a) x + 2 will be a factor iffi—2)
4
x
6x—
=
—
f(x) 3
f(—2)
=
(—2)
—
=
=
3
x
—
6x —4.
0
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.
0
—6
—4
—2
4
4
—2
—2
0
—2)1
1
2
The quotient is x
fix)
=
(x
+
2
2)(x
—
—
2x —2.
2x
—
2). These are the factors.
2 2x —2 had been factorable, then you would have listed
If x
its factors along with x + 2 in your answer.
—
Example 9
Factor completely: fix)
=
3
x
—
2
2x
—
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.
rs
The choices can be listed as rational numbers whose numerato
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
32
Module 8 Lesson 3
SenIor 3 Pre-Calculus Mathematics
Once you have listed the possible “a”
values, you can use the
Remainder Theorem and Factor Theore
m and/or synthetic
division to obtain the factors.
Note: When the leading coefficient is
1, then the possible “a”
values are simply the factor of the con
stant term.
3
2
fix)=x —2x —5x+6
•JE
C
L
Because the coefficient of x
3 is 1, the possible “a” values are the
factors of 6.
a
=
±1, ±2, ±3, ±6
Choose a = 1
If fla) = 0, then x
—
a is a factor
3
f(1)
=1 _2(1)2 —5(1)÷6
=1—2—5+6
=0
;.x—lisafactor
Use synthetic division to get the sec
ond factor:
1)1 —2 —5
6
1
—1
—6
1-—i. —6
0
F
r
quotient 2
is
—
x—
x
6
Notice that this is a quadratic that wil
l factor
—x--6=(x--3)(x÷2)
2
x
3
x
—
2
2x
—
5x +6= (x
—
1)(x
—
3)(x +2)
Li
[
4
:
33
Module 8, Lesson 3
Senior 3 Pro-Calculus Mathematics
Example 10
2
3 + 3x
Factor: 2x
—
8x
+
3.
Solution
Since the leading coefficient is 2 and the constant term is 3, the
possible a values are:
a
=
factors of 3
factors of 2
Check if a
=
=
±1, ± 3
±1,±2
=
±1, ± 3, ±
2
±
2
1
1
.1
)+3
)=2
2.
8(1
2
f(1
+
3
—
=2+3—8+3
=0
:.x—lisafactor
Use synthetic division to get a second factor:
1)2
3
2
2
—8
3
5—3
5—3
Secondfactor:
0
5x—3
2x
+
2
2 + 5x —3 = (2x 1)(x + 3)
This will factor: 2x
—
Factors are:
(x 1)(2x 1)(x
—
—
+ 3)
34
ModuleS, Lesson 3
SenIor 3 Pre-Calculus Mathematics
Assignment 3
1. Divide using long division and write in the form give
n by the
Division Algorithm.
3
2
a) —
3x+1÷x
x
—2
2
+
2
—
b) 3
3x—x
x
4÷x+2
c) 2
6
—
3
+
17x—6
1
x
÷3
6
x-x
2
2
+
3
—
—
d) 2
7—
3
x
4x
2
x
÷x
e) 2
4
x
—
3
+
8x+12÷
2
7
x+
x
1
[
L
2. Use synthetic division to find the quotient and the
remainder.
a)
b)
c)
d)
e)
x
—
3
7x+6÷x—2
3
—
4
x—4x
÷x—2
—
4
x
—
3
70x+20
2
÷x—
x
5
2
—
3
+
2
6x+3÷
x
5
2x—
x
1
4
+
3
—
2
7x—
4
x
6÷
x
2x+
3
3. Verifr whether or not x
+
2 is a factor of/tx)
=
3 +4x
x
2 —8.
4. Factor completely:
a) ±
2
f(
8
3
4x
x)=
x
+48
x
f(
—
4
—
3
+
b) 2
l8
x
xi
1
)=
2
7
-7
x
x
2
c) f(x) =
—27
2
f(
—
3
d) +
3x
x)1
=1
—
2
2x
x
3
5. When a polynomial fix) is divided by 2x + 1, the quot
ient is
2 x + 4 and the remainder is 3. Find fix).
x
—
C
SenIor 3 Pro-Calculus Mathematics
Module 8, Lesson 3
35
6.
—1
The volume of the rectangle prism is
-i-8x
3
V=3x
—
2
45x—5O.
Find the missing dimension.
7. The polynomial p(x) = 3
4x -i-bx
2 ÷ cx + 11 has a remainder of
—7 when divided by Cx +2) and a remainder of 14 when
divided byx —1.
•
SenIor 3 Pie-Calculus Mathematics
Module 8, Lesson 3
36
Notes
•
•Z
•••
U
U
0
TI
U
0
0
•••
C
{
Senior 3 Pre-Calculus Mathematj
Module 8, Lesson 3, Answer Key
15
Answer Key
Lesson 3
1. a)
2 + 4x +5
2x
x—2x+o3x+1
Answer
(x_2)(2x2+4x+5)+11
—4x
3
2x
2
•
—3x
2
4x
.
28x
5x+.1
fl
5x—1O
.
11
b)
—3
2
x
x+2÷2x2_3x
Answer
(x÷2)(x2_3)+2
2x2
—3x—4
rS-6
c)
2x 4x +3
2
16x .+ 17x —6
—
fj
3x
•
—
2x
—
—
2)(2X2
4x +3)
+17x
2
12x
-i-8x
2
—12x
9x—6
U
9x—6
d)
2
x
—
U
(3x
—4x
3
6x
2
U
ci
Answer
2x +3
_7
+ 3x
::
2
2
3x
—7
2
3x
—6
Answer
(x2
—
2)(2x + 3)—i
Senior 3 Pre-Calculus Mathematics
Module B, Lesson 3, Answer Key
16
swer
3x 4x + 12
2
2 +8x+12
—2x —7x
x+i)x4 8
e)
—
—
2 _4x+12)
(x+1)(x —3x
4
::::;::
x
8x
2
—4
+
x
4x
2
—4
—
12x+12
12x+12
0
2. a) 2)1
0
—7
6
2
4
—6
12—3
0
b) 2)3
3
c)
5)1
1
fl
2 + 2x —3
quotient: x
remainder: none
x
23
x
x+
ent:3
6
oti
12
2
qu
+
3
+
—4
—1
0
0
6
12
24
46
6
12
23
42
remainder:
—2
0
—70
20
5
15
75
25
3
15
5
45
+
r
quotient:
÷5
3x
15x
2
+
3
x
+
45
remainder: +
x—5
d) Before you do this question, 2x
1
X
form
t
valen
its equi
—
1 has to be changed mto
—
.
2
—5
6
3
2
quotient: 2x
.
remainder:
2-445
U
+
—
4x +4
5
2x_1,:
F
L
Senior 3 Pre-Calculus Mathematics
e)
Module 8, Lesson 3, Answer Key
..)4
2
4
—7
—6
17
quotient: 4x
2 + lOx +8
remainder: + 6
2x+3
.
1
4
10
8
6
3. You can do this in one of three ways:
a) Synthetic division
—2)1
1
4
0
—8
—2 —4
+8
2 —4
0
+-remainder
Because the remainder is 0, then x + 2 is a factor.
b) Long division
2 +2x—4
x
x+2)x3+4x2+0_8
2
3
x
i-2x
2x
+
2
0
fl
2x 2 +4x
—4x—8
—4x—8
0
Because the remainder is 0, then x ÷ 2 is a factor.
If f(—2) = 0, then x + 2 is a factor
2
f(
÷
3
—
x)
4
=x
x
8
U
f(—2) = (—2) + 4(2)2 —8
=—8+16—8
Since the remainder is 0, then x + 2 is a factor.
U
Module 8. Lesson 3, Answer Key
18
Senior 3 Pre-Calculus Mathematics
c) Use the Remainder Theorem
f(—2) gives the remainder
(—2)
f(—2)
..
+ 4(_2)2
—8
=—8+16—8
=0
x + 2 is a factor when the remainder is 0
8
x
=x
8
+4
x)
4x
2
f(
—
3
4. a) +
Possible “a values
=
=
factors of constant term
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.
8x +4x+48
)=x
2
f(x
—
3
1 _8(1)2 +4(1)÷48
3
f(1)=
=1—8+4÷48
x—lisnotafactor
=—45
...
8
(2)
+4
—8
(2)
)=2
i-4
2
f(2
=8—32+8+48
x—2isnotafactor
=32
..
3 _8(3)2 +4(3)÷48
3
f(3)=
27—72+12+48
:.x—3isnotafactor
=—5
=
f(4) =43_8(4)2+4(4) +48
64—128+16+ 48
x—4isafactor
=0
=
..
U
fl
Senior 3 Pre-Calculus Mathematics
Module 8, Lesson 3, Answer Key
19
Use synthetic division to find a second
factor
4)1 —8
4
48
U
4
2 —4x
x
—48
l2is the o:her factor which will factor
further
the factors are (x 4)(x
2 4x 12) or
(x —4) (x 6)(x + 2)
..
{J
fl
—16
—
—
V
—
—
—
b) f(x)
3
2x
1 7x
2 + 18x + 72
Possible a values = factors of constant term
factors of leadmg coefficient
= ±1, ±2, ±3, ±6, ±8,
±9, ±12,
±18, ± 24, ± 36, ± 72
=
—
—
,,
-
I
.
V
4
f(1
)=1 —2-1 _17(1)2÷1.8.1÷72
V
17
Q
=1—2—17+18+72
=72
x --1 is not a factor
f(—1) = (—i)
—
2(—1)
—
17(_1)2 +
18(—1) +72
=1÷2—17—18÷72
=40
x +1 is not a factor
1(2) = (2) —2- 2 17(2)2 + 18(2)+ 72
=16—16—68+36+72
=0
x +2 is not a factor
—
U
.U
0
f(—2) = (—2)
—
2(—2)
—
17(_2)2
+ 18(—2) +72
=16÷16—68—36+72
=0
:.x+2isafactor
o
3
....
...
,•
Senior 3 Pre-Calculus Mathematics
Module 8, Lesson 3, Answer Key
20
Use synthetic division to find another factor:
—2)1
1
—
2
4x
—2
—17
18
72
—2
8
18
—72
—4
—9
36
0
—
9x + 36 can still be a factor
)
6
4(3
9.3÷3
)3—
2
f(3
—
=27—36—27÷36
..x—3isafactor
=0
Use synthetic division again use x = 3 and
4x 36.
9x+
—
3
x
—
2
3)1
1
—4
—9
36
3
—3
—36
—1
0
—12
2 x 12 which factors into (x
quotient is x
thefactorsare(x+2)(x—3)(x+3)(x—4)
—
—
—
4)(x + 3)
F
3
c) f(x)x —27
Possible “a” factors: ±1, ±3, ±9, ±27
27
)=3
3
f(3
—
=27—27= 0
x —3 is a factor
Do synthetic division:
3)100
-27
9
27
139
0
3
JE
2 + 3x + 9— does not factor further.
Other factor is x
(x + 3x + 9)
factors are (x 3) 2
—
[
C
q
r
Senior 3 Pre-Calculus Mathem
atics
Module 8, Lesson 3, Answer Key
21
d) +
2
f
—
3
(x
x)
—
3
4
(4
1
x
x
)
Possible a values
=
=
f(1)
=
3(4(1)
-
4(1)2
factors of constant term
factors of leading coefficient
+1
=+1+_+_
±1,±2,±4
‘2’4
—
+ 11)
=3(4—4+1—1)
fl
=
0
:.
x —1 is a factor
Use synthetic division to ge
t another factor.
1)4 —4 1 —1
U
fl.
4010
The three factors are 3(4
2
x + 1)(x
5. f(x)=(2x+1)(x2_x+
4)+3
11
—
1)
=
—
3
+
2
—
8
x
2
x
+
2
4
+
x
+
x
x
3
=
—
3
÷
2
72
x+
x
x
7
U
6. The volume of the rectan
gular prism is
V 3
3x 2
+
8
—
45x—
x
50.
Use synthetic division with
V and x + 1.
—1)3
8
—45
—50
—3
—5
+50
5
—50
0
3
U
..
The quotient is 3x
2 + 5x —50 which factors int
o
(3x—10)(x+5)
The missing term is 3x 10.
—
U
3.
.
.,,.,
.
Module 8, Lesson 3, Answer Key
22
f(—2)
=
4(—2)
+
+
Senior 3 PrealcuIUs Mathematics
C
c(—2)+11 which equals —7
7
_32+4b—2X11
(1)
4b—2c=14
f(1) = 4(1)
+
2
b(1)
+
c(1) + 11 which equals 14
144+b+c+11
b+c+1514
D
fl
(2)
b+c—1
2c =14
b+c=—l
4b
—
4b—2c=14
2b+2c—2
6b=12
2
Eqn(2)X
U
3
c=—
+X+
3
p(x)=4x
3
2
11
2X
t
t
r