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A Teaching- Learning Package in Mathematics 10

Teacher: Maryland V. Pasilan

Method: Five A’s

TLP 5: Remainder and Factor Theorem

Objectives: Through the discussion and examples, the student will be able to:
1. Evaluate a polynomial function,
2. Use the remainder and factor theorem to find the remainder when a polynomial is divided
by a binomial, and
3. Factor polynomials completely using remainder theorem, and factor theorem.

A. ACTIVATION
Awareness
Questions are presented to the class.
 What are the essential method to evaluate a polynomial?
 How can we say that a binomial is a factor or not?
 What is the use of the substitution in evaluating a polynomial?
 Can evaluating a polynomial can identify the following factors of a polynomial?

Activity
Revealing the Message
Class will be divided into two groups. Each group is given their function to solve. Example is given.
Directions: Find the value of y using the given value of x to reveal the message. Choose the correct answer in
the box with corresponding letter.
Group A. y = x3+x2+x+3 Group B. y = x4-4x3-7x2+2x+18
Example:
Group A
IF x is = -2 what is the value of y?
Using the equation above, y = x3 + x2+x+3, substitute -2 to the x. So, we have:
y = x3 + x2+x+3 Given
y = (-2)3 + (-2)2+-2+3 Substitute -2 to x
y = -8 + 4 -2+3 Simplify. (-2)3 = -8 ; (-2)2 = 4
y = -4 + 1 -8 + 4 = -4 ; -2 + 3 = 1
y = -3
Therefore, the value of y if the x is -2 is -3.
Look for the answer in the box. The answer is corresponding with a letter.
Group B
If x = -2, what is the value of y using the equation y = x4-4x3-7x2+2x+18?
Same step with the above, use the equation and substitute -2 to x.
y = x4-4x3-7x+2x+18 Given
4 3 2
y = (-2) -4(-2) -7(-2) +2(-2)+18 Substitute -2 to x.
y = 16 + 32 -28 -4 + 18 Simplify. (-2)4 = 16 ; -4 (-2)3 = -32 ; -7(-2)2 = -28 ; 2(-2) = -4
y = 34 Simplify.
Therefore, the value of y if the x is -2 is 34
Look for the answer in the box. The answer is corresponding with a letter.

A B
x -2 -1 0 1 2 -2 -1 0 1 5
1 2 3 4 5 6 7 8 9 10
y -3 ___ ___ ___ ___ 34 ___ ___ ___ ____
1 2 3 4 5 6 7 8 9 10
Message C ___ ___ ___ ___ _S__ ___ ___ ___ ___

After, the two groups will combined the words they made to form the message.
Questions:
How did you get the answers?

What pattern you have observed ?

What method you used to solve the equation?
What is the message?

B. BUILDING – UP
Analysis
The activity above will help you to understand the use of reminder and factor Theorem. We evaluate
equation above using the given values of x. In remainder theorem we apply this.
The Remainder Theorem
When computing in finding the remainder when the polynomial p(x) is divided by x-c can be easily
determined by simply evaluating p(x) for x = c. In other words, simply find p(c). This is the essence of the
Remainder Theorem.
The Remainder Theorem
If a polynomial p(x) is divided by x-c, then the remainder is equal to p(c).
Examples:
In the division of polynomials, we have encountered the linear polynomial x – c if we use synthetic division. In
this matter, it is very use to find the remainder of a given polynomial. To find understand let us study the given
examples.
Example 1
Find the remainder when P(x) = 5x3+2x-3 is divided by x+2.
Solution:
 Find the value of c from the divisor x-c.
x+2 = x-(2) = x - 2 ; hence c = -2
Evaluate P(c) by substituting c in the dividend.
P(x) = 5x3+2x-3 Given
P(-2) = 5(-2)3 + 2(-2)-3 Substitute 2 to x.
= 5(-8)-4-3 9(-2)3 = -8 ; 2(-2) = -4
= -40-4-3 Simplify.
P(-2) = -47
Therefore, the remainder 5x3+2x-3 is divided by x+2 is -47.
To Verify, we use synthetic division

Example 2
Find the reminder when the given p(x) is divided by the given divisor x-c.
p(x)= x87+2x48+7 is divided to x+1
Solution:
x = x and c = 2.
Substitute the values to x-c = x – 2, thus c = -1.
c = -1
p(x)= x87+2x48+7 Given.
p(-1) =(-1)87+2(-1)48+7 Substitute -1 to x.
= -1+2(1)+7 Simplify.
=8
The remainder when x87+2x48+7 is divided by (x+1) is 8.

Example 3
Find the reminder when the given p(x) is divided by the given divisor x-c.
P(x) = 2x7 – 3x5 +4x3 -5x+3 is divide to x-2
Solution:
x = x, c = -2
Substitute the values to x –c, x – (-2) = x + 2, therefore c = 2.
c=2
P(x) = 2x7 – 3x5 +4x3 -5x+3 Given.
p(2) = 2(2)7 -3(2)5 +4(2)3 – 5(2) +3 Substitute 2 to x.
= 256-96+32-10+3 Simplify.
p (2) = 185
The remainder when (2x7 – 3x5 +4x3 -5x+3) is divided by (x-2) is 185.

Example 4
When x5-3x4-2x3+4x2-12x+3 is divided by x+1, What is the remainder?
Solution:
x = x, c = 1
Substitute the values to x –c = x – 1, therefore c =- 1.
c=-1
p(x) = x5-3x4-2x3+4x2-12x+3 Given
P(-1) = (-1)5 -3(-1)4 -2(-1)2-12(-1)+3 Substitute -1 to x.
= -1-3+2+4-12+3 Simplify.
= -7
The remainder when x5-3x4-2x3+4x2-12x+3 is divided by (x+1) is -7.
To Verify, we use synthetic division

Therefore, using the Remainder Theorem we can identify that the given binomial is factor or not.

Try This # 1!
Find the reminder when the given p(x) is divided by the given divisor x-c.
1. a4-14a2+5a-3 ; a+4 2. 5a2-2a+5; a-3
[Students are tasked to this on their own. After, students will be called to explain their answers
with the guidance of the teacher.]

The factor Theorem and its converse.

The use of Remainder Theorem gives a hint to the factorability of a divisor in the form x-c. The
binomial x-c is a factor of p(x) if p|(x) can be divided exactly by x-c. That is,
p (x)
=q ( x ) +0.
x−c
This relation implies that if the Remainder Theorem is applied, p(c) = 0. This observation is expressed in
the next theorem.
The Factor Theorem and Its Converse
If p(c) = 0, then x-c is a factor of polynomial p(x).
Conversely, if x-c is factor of polynomial p(x), then p(c) =0.

Example 1
Determine that x-2 is factor of x4+2x3-37x2+82x-48.
Solution:
Solve for p(2).
Since x-c = x-(-2) = x+2, so, c = 2
p(x) = x4+2x3-37x2+82x-48 Given.
 p(2) = (2)4+2(2)3-37(2)2 +82(2) – 48 Substitute 2 to x.
= 16 +2(8) -37(4) +164 -48 Then, simplify.
= 16+16-148+164-48
p(2) = 0.
Since P(2) = 0, therefore x-2 is factor of x4+2x3-37x2+82x-48.
Example 2
Determine whether x-3 is factor x4-6x2-15x+6.
Solution:
Solve for p(3).
Since x-c = x-(-3) = x+3, so, c = 3
p(x) = x4-6x2-15x+6 Given.
p(3) = (3)4-6(3)2-15(3)+6 Substitute 3 to x.
= 81 -6(9) – 45 +6 Then, simplify.
= 81 -54-45+6
P(3) = -12
Since p(3) ≠ (not equal to) 0, therefore, x-3 is NOT a factor x4-6x2-15x+6.

Practice!
Try This #2!
Determine whether x+4 is a factor of 2x4+5x2-8x+16.
[Students are tasked to this on their own. After, students will be called to explain their answers with
the guidance of the teacher.]

Example 3
Determine which of the following binomials are factors of p(x) = 2x3-7x2-5x+4.
a. x+1 b. x-2
1
c. x-4 c. x-
2
Solution:
a. x+1
x-c=x+1
= x -1
c = -1
Find the remainder by applying he Remainder Theorem.
p(x) = 2x3-7x2-5x+4 Given.
P(-1) =2(-1)3-7(-1)2-5(-1)+4 Substitute -1 to x.
= -2-7+5+4 Then, simplify.
=0
By the Factor Theorem, since p(-1)= 0, therefore ,x+1 is factor of p(x).
b. x -2
x-c = x-2
= x – (-2)
=x+2
c=2
Find the remainder.
p(x) = 2x3-7x2-5x+4 Given.
p(2) = 2(2)2-7(2)2-5(2)+4 Substitute 2 to x.
= 16-28-10+4 Then, simplify.
= -18
Since p(2) = -18 ≠ 0, x-2 is not a factor of p(x).
c. x -4
x-c = x-4
= x – (-4)
=x+4
c=4
p(x) = 2x -7x2-5x+4
3
Given.
p(4) = 2(4)3 – 7(4)2 -5(4) +4 Substitute 4 to x.
= 128 -112-20+4 Then, simplify.
=0
Since p(4) = 0, x-4 is factor of p(x).
1
d. x-
2
1
x-c = x-
2
1
= x – (- )
2
1
=x+
2
1
c=
2
p(x) = 2x3-7x2-5x+4 Given.

() () () ()
3 2
1 1 1 1 1
p =2 -7 −5 +4 Substitute to x.
2 2 2 2 2

= 2( ) - 7( ) - ( ) + 4
1 1 5
Then, Simplify.
8 4 2
1 7 10 16
= − − +
4 4 4 4
=0

Since p ( 12 ) = 0, x- 12 is factor of p(x).

Example 4
Determine if x-c is a factor of p(x).
a. p(x)=x3+x2-7x-3; x+3
b. p(x)=5x3+4x2-31x+6; x-2
Solution:
The Factor Theorem implies that if p(c)=0, then p(x) is divisible by x-c.
a. p(x)=x3+x2-7x-3; x+3
x-c = x+3
= x -3
c = -3
p(x)=x3+x2-7x-3 Given.
p(-3) = (-3)3 +(-3)3 -7(-3)-3 Substitute -3 to x.
= -27+9+21-3 Then, simplify.
=0
Since p(-3) = 0, x+3 is factor of p(x). Therefore, p(x) is divisible by x+3.
b. p(x)=5x3+4x2-31x+6; x-2
x-c = x - 2
= x –(- 2)
= x+ 2
c=2
p(x)=5x3+4x2-31x+6 Given.
p(2) = 5(2)3+4(2)2-31(2)+6 Substitute 2 to x.
= 40 +16-62+6 Then, simplify.
=0
Since p(2) = 0, x-2 is factor of p(x). Thus, p(x) is divisible by x-2.

Note that the statement “x-c is factor p(x)” implies that p(x) is divisible by x-c.
The statement “x-c is factor of p(x)” also implies that c is a root of p(x) = 0.

Practice!
Try This #3!
Check if x-c is a factor of the p(x).
1. p(x) = x3-2x2+x-12; x-2 2.p(x) = 2x4-x-3; x+1
3. p(x) = x5-3x4+8x+2; x-2
[Students are tasked to this on their own. After, students will be called to explain their answers with the
guidance of the teacher.]

Abstraction
Using a Venn diagram, compare and contrast factor and remainder theorem.

C. CLOSURE
Application
A. Find the reminder when the given p(x) is divided by the given divisor x-c. Identify whether it is a factor or
not.
1. p(x) = 5x2+2x-4; x-1 2. p(x) = 3x4-5x3+x2-8x+3; x+1

3. p(x) = x5-6x+9;x-2 4. P (x) = 2x3-x2+x;x+2

5. p(x ) = x3 + 6x2 +11x + 6; x +1 6. p(x) = x3 – 7x + 6; x +3

7. p(x) = x3 – 3x2 – 4x + 12; x + 13 8. p(x) = x3 + x2 – 12x; x + 4

9. p(x) = 2x3 – 7x2 + 7x – 2; x + 2 10. p(x) = 9x3 – 7x + ; x - 3

B. Choose form the boxes that contains the factors of the given polynomial. Encircle the factor or factor of the
given polynomial. Show your solutions.

1. x3-5x2-2x+24 x+1 x-1 x+3 x-3

2. -5x3+14x+12 x+2 x-2 x+3 x-3
3. x3+2x2-9x-18 x+2 x-2 x+3 x-3
4. x3-3x2-x+3 x+1 x-1 x+3 x-3
5. x3+4x2+3x-2 x+1 x-1 x+2 x-2
6. x3 – 6x2 + 11x -6 x+2 x-2 x+3 x- 3
7. x4 + 2x3- 4x2 + 9 x+1 x -1 x+3 x-3