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Mathematics
Quarter 1 – Module 9:
Remainder Theorem, Factor Theorem
and the Rational Root Theorem

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Mathematics – Grade 10
Quarter 1 – Module 9: Remainder Theorem, Factor Theorem and the Rational Root
Theorem

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Regional Director: Gilbert T. Sadsad

Assistant Regional Director: Jessie L. Amin

Development Team of the Module

Writer: Evelyn P. Balang

Editors: Salvacion B. Felices
Jon Jon R. Monte
Shiela L. Guevara
Noel A. Lozano
Alfie T. Gascon
Jinky A. Villareal
Reviewer: Jinky A. Villareal
Illustrator: Ryan B. Cerillo
Layout Artist: Anthony C. Vista

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 Lesson
REMAINDER THEOREM, FACTOR THEOREM
AND RATIONAL ROOT THEOREM

You want to know?

In this lesson, you will learn a new method of finding the remainder when a
polynomial is divided by x – r. You will also learn a method of determining
whether or not x-r is a factor of a given polynomial and also let you find all
the possible zeros of a polynomial equation. This module will give you an
idea on how it is done through activities.
.

The important skills that you will learn in this lesson are:

1. proving the remainder theorem, factor theorem and the

rational root theorem, specifically determining whether the
binomial x-r is a factor of the given polynomial; and

2. finding the possible zeros of a polynomial equation.

The following are terms that you must remember from this point on
1. Remainder- An amount left over after division ( this is a result when
the first number does not divide exactly by the other.
2. Factor- a number or algebraic expression that divides another
number or expression exactly / with no remainder.
3. Theorem – a theorem is a statement that can be demonstrated to
be true by accepted mathematical operations and arguments.
4. Remainder Theorem – If the polynomial P(x) is divided by (x – r),
the remainder R is a constant and is equal to P(r).
R = P(r)
5. Factor Theorem- states that (x-a) is a factor of the polynomial P(x)
if and only if P(a) = 0.
6. Rational Root Theorem- describes a relationship between the roots
of a polynomial and its coefficients specifically the nature of any
rational roots the polynomial might possess

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 PRE-TEST

Instruction: Read each item carefully then choose the letter of your answer among the options.
Write only the letter on your answer sheet.

1. What is the remainder when P(x) = x3 + 5x -7 is divided by x + 3?

A. 49 B. -49 C. 5 D. -5

2. Use the remainder the Remainder Theorem to find the remainder when
P(x)= 3x3 + 4x2 -5x +6 is divided by x+3.

A. 36 B. 24 C. -24 D. 66

3. Which example correctly illustrates the factor Theorem?

A. f(x) = x3 – x2 + x – 1 ; f(1) = 0, so x- 1 is a factor of f (x).
B. f(x) = x3 – x2 + x – 1 ; f(1) = 0, so x +1 is a factor of f (x).
C. f(x) = x3 – x2 + x – 1 ; f(2) = 5, so x - 2 is a factor of f (x).
D. f(x) = x3 – x2 + x – 1 ; x- 1 is a factor of f(x), so f(-1) =0.

4. If P(-2) = 0, which of the following statements is true about P(x) ?

A. x + 2 is a factor of P(x)
B. 2 is a root of P(x) = 0
C. P(x) = 0, has 2 negative roots
D. P(0) = -2

5. Which of the following is a zero of x4 – 3x2 + 2 = 0?

A. -1 B. 2 C. -2 D. 3

How did you find the test?

Don’t worry class if you got a low
score. I’m sure you will learn
more as you go through this
module. Just have fun learning
Math.

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 LEARNING ACTIVITIES

In the previous lesson, you learned how to divide polynomials

using long division and synthetic division which is helpful in our lesson
today. To go through this module, you will learn the 3 theorems that will
help you determine if the polynomial is a factor of another polynomial
and to determine the possible roots of the polynomial

Do you remember doing division in Arithmetic?

15 ÷ 6 = 2 R 3

“15 divided by 6 equals 2 with a remainder of 3

Each part of the division has names:

Dividend 7÷2=3 R1 Remainder

Divisor Quotients

Which can be rewritten as a sum like this:

Dividend 7=2×3+1 Remainder

Divisor Quotient

We can also divide polynomials.

Dividend f(x) = d(x) · q(x) + r(x) Remainder

Divisor Quotient

In division of polynomials, the quotient and the remainder can be obtained by long division.
However, when the divisor is in the form x – a or mx – n, then this is a simpler method to find the
remainder.

Consider when a polynomial f(x) is divided by x – a. By division algorithm, we have:

f(x) = (x – a) × Q(x) + R, (1)
where Q(x) and R are the quotient and the remainder, respectively.

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 By substituting x = a into (1), we have
f(a) = (a – a) × Q(a) + R
= 0 × Q(a) + R = R
Thus, we have following remainder theorem.

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

Example 1

Let P(x) = 3x3 + 4x2 -5x + 6. Use the remainder Theorem to evaluate P(-3).

Solution:
The value of P(-3) is the remainder obtained when P(x) is divided by x – (-
3). Use synthetic division to divide P(x) by x + 3.

-3 3 4 -5 6
-9 15 -30

3 – 5 10 - 24 ← 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟

The remainder, - 24, is the value of P(-3). Thus, P(-3) = -24. This result can be
verified by evaluating P(-3) directly. Using P(x) = 3x3 + 4x2 – 5x + 6, we obtain
P(-3) = 3(-3)3 + 4(-3)2 – 5(-3) + 6 = -81 + 36 + 15 + 6 = -24

Example 2.

If P(x) = x4 – 10x3 + 2x2 – 8x + 5, find P(10).

Solution:
When P(x) is divided by x – 10, the remainder is P(10). Performing
synthetic division:
10 1 -10 2 -8 5
10 0 20 120

1 0 2 12 125
By the Remainder Theorem, P(10) = 125.

Check: Perform direct substitution.

P(10) = (10)4 – 10(10)3 + 2(10)2 – 8(10) + 5

125 = 10 000 – 10 000 + 200 - 80 + 5
125 = 125

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Example 3

The expression x3 – kx2 + 5x + 3 leaves a remainder of -11 when divided by x –

2. Find the value of k.
Solution:
Let P(x) = x3 – kx2 + 5x + 3.

By the Remainder Theorem,

P(2) = -11
(2)3 – k(2)2 + 5(2) + 3 = -11 Replace x with 2
8 – 4k + 10 + 3 = -11 Simplify
-4k + 21 = -11 Add
-4k = -32 Subtract 21 from both sides
k=8 Divide both sides by 4

Did you understand the remainder theorem?

How do we determine the remainder of a polynomial using the remainder
theorem?

Find the remainder when P(x) = x3 + 5x -7 is divided by x – 2.

Were you able to get the answer correctly? Congrats!

I think you are ready for the next activity. Let’s proceed…

Factor Theorem

When f(x) = 2x3 + 5x2 + x – 2 is divided by x + 2, the remainder obtained is 0.

By division algorithm, we have

f(x) = (x + 2) × Q(x) + 0,
= (x + 2) × Q(x) where Q(x) is the quotient. Remainder = 0

This means that x + 2 is a factor of f(x), that is f(x) is divisible by x + 2.

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The above example illustrates the following factor theorem.

Factor Theorem
For a polynomial f(x), if f(a) = 0, then x – a is a factor of f(x).
Conversely, if x – a is a factor of a polynomial f(x), then f(a) = 0.

Example 1
Determine whether x – 4 is a factor of each polynomial.
a. x3 – 2x2 – 11x + 12 b. 2x3 + x2 – 2x – 15
Solution:
a. Let P(x) = x3 – 2x2 – 11x + 12
P(4) = (4)3 – 2 (4)2 – 11(4) + 12
= 64 – 32 – 44 + 12
=0
∴ x – 4 is a factor.

b. Let P(x) = 2x3 + x2 – 2x – 15

P(4) = 2(4)3 + (4)2 – 2(4) – 15
= 128 + 16 – 8 – 15
= 121
Since P(4) ≠ 0, x – 4 is not a factor of 2x3 + x2 – 2x – 15.

Example 2
Show that (x – 1) is a factor of 3x3 – 8x2 + 3x + 2.
Solution:
Using the Factor Theorem, we have:
P(1) = 3(1)3 – 8(1)2 + 3(1) + 2
=3–8+3+2
=0
Since P(1) = 0, then x – 1 is a factor of 3x2 – 8x2 + 3x + 2.
Example 3
Find the value of k for which the binomial (x + 4) is a factor of x4 + kx3 – 4x2.
Solution:
If (x + 4) is a factor of P(x) = x4 + kx3 – 4x2, we know from the Factor
Theorem that P(-4) = 0.
P(-4) = (4)4 + k(-4)3 – 4(-4)2 = 0
256 – 64k – 64 = 0
64k = 192
64 64
k=3

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 Find k so that x + 3 is a factor of
a. 2x3 + kx2 – 8x + 3 b. x3 + 3x2 – kx – 3

How did you find the lesson? Were you able to determine if the given
polynomial is factor of the other? I know you were…
Are you still okay? I hope you are, because we still have one more
activity which is more interesting.

To help you understand the next theorem, consider the polynomial

P(x) = (x + 3) (x – 1) (x – 5) Factored Form

= x3 – 3x2 – 13x + 15 Expanded Form

From the factored form we see that the roots of P(x) are 1, 5, and -3. When the polynomial
is expanded, the constant 15 is obtained by multiplying (-1) (-5) (3).
This means that the roots of the polynomial are all factors of the constant term. The
following coefficients then, generalizes this observation.

Rational Roots Theorem

Consider the polynomial equation P(x) = a n xn + an – 1 xn – 1 + ⋯ + a1 x + ao = 0 (where n ≥

𝑝
1, an ≠ 0) has integer coefficients. Let 𝑞 be a rational number, where p and q have no common
𝑝
factors other than 1. If 𝑞
is a root of P(x) = 0, then p is a factor of the constant coefficient a 0 and
q is a factor of leading coefficient a n.

𝑝
Proof: If 𝑞 is a rational root, in lowest terms, of P(x) = 0, then we haves

an (𝑝𝑞) n + an – 1 (𝑝𝑞)n – 1 + ⋯ + a1 (𝑝𝑞) + a0 = 0

an pn + an – 1 pn – 1 q + ⋯ + a1 pqn – 1 + a0 qn = 0 Multiply by qn.

p (an pn – 1 + an – 1 pn – 2 q + ⋯ + a1 qn – 1) = -a0 qn Add by -a0 qn to both sides and factor

LHS.

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 𝑝
Now p is a factor of the left side, so it must be a factor of the right as well. Since 𝑞
is in
lowest terms, p and q have no common factors in common, and so p must be a factor of a0. A
similar proof shows that q is a factor of an.

Rational Root Theorem

𝑝
The possible rational roots are formed as ± 𝑞 , where “p” is a factor of a constant
(end) term and “q” is a factor of the leading term

Example 1.
Find the possible rational roots of the equation 2x3 – x2 – 9x – 4 = 0
Solution: First we list the factors of a 0 ( the constant term ) and the factors of a n the
numerical coefficient of the first term)
Factors of p or a0= -4 : ±1 , ±2 , ±4
Factors of q or a3 = 2 : ±1, ±2

1 1 2 2 4 4
Divide each factor of p by q : ±1 , ± , ± , ± ,± ,±
2 1 2 1 2
Possible roots = ±1 , ±½ , ±2 , ±4

Example 2.
List all the possible rational zeros of f(x) = 3x3 + 41x2 + 121x – 45.
Solution:
𝑝
The possible rational zeros are 𝑞, where p is a factor of 45 and q is a factor of
3.
The possible values of p: ±1, ±3, ±5, ±9, ±15, ±45
The possible values of q: ±1, ±3
1 1 3 3 5 5 9 9
Thus, the possible rational zeros are: ± 1, ± 3 , ± 1 , ± 3 , ± 1 , ± 3 , ± 1 , ± 3 ,
15 15 45 45
± 1
, ± 3
, ± 1 , and ± 3
1 5
Equals: ±1, ± , ±3, ±5, ± , ±9, ±15, 𝑎𝑛𝑑 ± 45
3 3

Congrats my dear students, you are done with the first part of this
module!
At this time, you are task to answer the following activities to enhance
your skills and understanding of the lessons discussed in this module.

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 PRACTICE EXERCISES

PRACTICE TASK 1

Find the remainder for each of the following using the remainder theorem.
1. ( x2 – 3x + 4 ) ÷ ( x – 1 )
2. ( 2x3 + 5x2 + x – 2) ÷ ( 2x + 1)
3. ( x4-10x3 + 2x2 – 8x + 5) ÷ (x – 10)
4. (x3 + 4x2 + 3x – 2) ÷ (x – 3)
5. (3x2- 5x + 4) ÷ ( x-1)

PRACTICE TASK 2

Draw a star ( ) if the first polynomial is a factor of the second if not draw a triangle
( ).

1. X–1 ; x2 + 2x + 5
2. a+1 ; 2a3 + 5a2 - 3
3. y+3 ; 2y3 + y2 – 13y + 6
4. X–1 ; x3 – x – 2
5. x–3 ; -4x3 + 5x2 + 8

PRACTICE TASK 3

List all the possible zeros of each polynomial.

1. Y3 + 9y2 – 44 = 0
2. X4 - 3x2 + 2 = 0;
3. 2x2 – 5x – 3 = 0
4. X3 – 2x2 – 11x + 12 =0
5. X4 – 2x2 – 16x – 15=0

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 POST TEST

Instruction: Read each item carefully then choose the letter of your answer among the options.
Write only the letter on your answer sheet.

1. If P(x) is divided by x – a, then

A. P(a) = divisor C. P(a) = Quotient

B. P(a) = dividend D. P(a) = remainder
2. What is the remainder when P(x) = x3 + 5x -7 is divided by x + 3?

A. 49 B. -49 C. 5 D. -5

3. Find k so that x + 3 is a factor of 2x3 + kx2 – 8x + 3.

A. 2 B. -2 C. 3 D. -3

4. When 10x3 + mx2 – x + 10 is divided by 5x- 3 the remainder is 4. Find the possible
value of m.

A. -19 B. -21 C. 19 D. 21

5, How many possible roots this polynomial P(x) = 3x4 + 5x2 – 20 have?

A. 6 B. 12 C. 18 D. 24

ADDITIONAL ACTIVITIES

1. Show that P(x)= x3- 6x2 + 11x – 6 is divisible by x-1.

2. Use the Remainder Theorem to determine whether x = -4 is a solution of x6 +
5x5 + 5x4 + 5x3 + 2x2 – 10x – 8.
3. List all the possible roots of 3x5 – x4 + 6x3 – 2x2 + 8x – 5 = 0

Thank you learners…

Stay safe and healthy!

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 ANSWER KEY

PRE- TEST
1. B
2. C
3. A
4. A
5. A

Practice Task 1
1. 2
2. -3/2
3. 125
4. 70
5 2

Practice Task 2
1.
2.
3.
4.
5.

Practice Task 3
1. ±1, ±2, ±4, ±8, ±11, ±22
2. ±1, ±2
1 3
3. ±1, ± 2 , ±3, ± 2
4. ±1, ±2, ±3, ±4, ±6, ±12
5. ±1, ±3, ±5, ±15

POST- TEST
1. D
2. B
3. C
4. B
5. D

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BIBLIOGRAPHY
A. References:
1. Mathematics Learner’s Module for Grade 10 ( DEPED )
2. Math for the 21st Century Learners for Grade 10
3. E- Math for Grade 10

B. Website Links as References and Sources of Learning Activities.

➢ https:// www.britannica. Com
➢ https://www.mathsisfun.com
➢ http//www.purplemath.com

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