The document discusses polynomial division and related concepts. It defines polynomial expressions and covers the algorithms for long division and synthetic division of polynomials. It also introduces the factor theorem and remainder theorem, which relate the factors and remainder of a polynomial division to evaluating the polynomial at specific values. The document provides examples of applying these techniques to divide polynomials and determine factors and remainders.

1. DIVISION OF
POLYNOMIALS
A. Polynomial Expressions
B. Division of Integers
C. Algorithm in Division of Polynomials
D. Long Division
E. Synthetic Division
F. Factor and Remainder Theorem

2. WHATISAPOLYNOMIALEXPRESSION?
A polynomial expression P(x) is an algebraic expression of the
form anxn + an-1 xn-1 + … + a1x + a0, an ≠ 0 where the
nonnegative integer n is called the degree of the polynomials
and coefficients a0, a1, …, an are real numbers.
POLYNOMIALEXPRESSION NOTAPOLYNOMIALEXPRESSION
3𝑥 + 2 2𝑎(𝑎−3 + 𝑎2 − 1)
(6𝑥 − 2) 𝑥 + 1 2(𝑥 − 1) 5𝑥−2
− 4𝑥
1
2 − 3
𝑥3 − 5𝑥2 + 𝑥 + 3
1
𝑥

3. −28
DIVISIONOF
INTEGERS
• The quotient of two integers with the SAME SIGN is a
POSITIVE REAL NUMBER. ( +/+ or -/-)
• The quotient of two integers with DIFFERENT SIGNS
is a NEGATIVE REAL NUMBER. ( +/- or -/+)
1. 14)350
LET’S TRY! Solve the following integers.
2
70
5
−70
0
2. 11)143
1
−11
33
3
−33
0
3. 42)168
4
−168
0

4. THEDIVISIONALGORITHM

5. DIVISIONOFPOLYNOMIALSUSING
LONG DIVISION
Step 1: Divide the first
term of the dividend by
first term of the divisor to
get first term of the
quotient.
Step 2: Take the term
found in step 2 and
multiply it times the
divisor.
Step 3: Subtract this from
the line above.
Step 4: Repeat until done.
Step 5: Write the answer.
2𝑥 + 3)8𝑥3 + 0𝑥2 + 0𝑥 + 27
Divide 8𝑥3
+ 27 𝑏𝑦 (2𝑥 + 3).
4𝑥2
−8𝑥3
+12𝑥2
−12𝑥2
−6𝑥
+0𝑥
−12𝑥2
− 18𝑥
−18𝑥 + 27
−9
−18𝑥 − 27
0

6. DIVISIONOFPOLYNOMIALSUSING
LONG DIVISION
Step 1: Divide the first
term of the dividend by
first term of the divisor to
get first term of the
quotient.
Step 2: Take the term
found in step 2 and
multiply it times the
divisor.
Step 3: Subtract this from
the line above.
Step 4: Repeat until done.
Step 5: Write the answer.
𝑥 − 3)2𝑥2 − 𝑥 − 24
Divide 2𝑥2
− 𝑥 − 24 𝑏𝑦 (𝑥 − 3).
−2𝑥2
−6𝑥
5𝑥
2𝑥
−24
−5𝑥 − 15
+5
9
*In writing the remainder, the remainder itself should be the numerator
and the divisor will be the denominator.
𝟐𝒙 + 𝟓 +
𝟗
𝒙 − 𝟑

7. Divide 2𝑥3
− 3𝑥2
− 5𝑥 − 12 by (𝑥 − 3).
𝑥 − 3 )2𝑥3 − 3𝑥2 − 5𝑥 − 12
2𝑥2
−2𝑥3
− 6𝑥2
3𝑥2
− 5𝑥
+ 3𝑥
−3𝑥2
− 9𝑥
4𝑥 − 12
+ 4
−4𝑥 − 12
0
Long Division
|
|
V
Synthetic Division
STAGE 1: LONG DIVISION

8. Divide 2𝑥3
− 3𝑥2
− 5𝑥 − 12 by (𝑥 − 3).
−3 )2 − 3 − 5 − 12
2 + 3 + 4
−2 − 6
3 − 5
−3 − 9
4 − 12
−4 − 12
0
STAGE 2: VARIABLES SUPPRESSED

9. Divide 2𝑥3
− 3𝑥2
− 5𝑥 − 12 by (𝑥 − 3).
STAGE 3: COLLAPSED VERTICALLY
-3 2 -3 -5 -12
-6 -9 -12
2 3 4 0
Dividend
Quotient,
remainder

10. Divide 2𝑥3
− 3𝑥2
− 5𝑥 − 12 by (𝑥 − 3).
STAGE 4: SYNTHETIC DIVISION
3 2 -3 -5 -12
6 9 12
2 3 4 0
Dividend
Quotient, remainder
• The number standing for the divisor x – k is now k, its zero.
• Changing the signs in the second line allows us to add rather than subtract.

11. • Synthetic division is a method used to perform the
division operation on polynomials when the divisor is a
linear factor.
DIVISIONOFPOLYNOMIALSUSING
SYNTHETIC DIVISION

12. DIVISIONOFPOLYNOMIALSUSING
SYNTHETIC DIVISION
Step 1: Check whether the polynomial is in the standard form.
Step 2: Write the coefficients in the dividend's place and write the zero of the
linear factor in the divisor's place.
Step 3: Bring the first coefficient down.
Step 4: Multiply it with the divisor and write it below the next coefficient.
Step 5: Add them and write the value below.
Step 6: Repeat the previous 2 steps until you reach the last term.
Step 7: Separate the last term thus obtained which is the remainder.
Step 8: Now group the coefficients with the variables to get the quotient.

13. DIVISIONOFPOLYNOMIALS USING
SYNTHETIC DIVISION
Step 1: Check whether the
polynomial is in the standard form.
Step 2: Write the coefficients in the
dividend's place and write the zero of
the linear factor in the divisor's place.
Step 3: Bring the first coefficient
down.
Step 4: Multiply it with the divisor and
write it below the next coefficient.
Step 5: Add them and write the value
below.
Step 6: Repeat the previous 2 steps
until you reach the last term.
Step 7: Separate the last term thus
obtained which is the remainder.
Step 8: Now group the coefficients
with the variables to get the quotient.
Divide 2𝑥2 − 𝑥 − 24 𝑏𝑦 (𝑥 − 3).
2 − 1 − 24
To know the zero of the divisor or linear factor, equate 𝒙 − 𝟑.
𝑥 − 3
𝑥 − 3 = 0
𝑥 − 3 + 3 = 0 + 3 Addition Property of Equality
𝒙 = 𝟑
3
2
6
5
15
−9

14. DIVISIONOFPOLYNOMIALS USING
SYNTHETIC DIVISION
Step 1: Check whether the
polynomial is in the standard form.
Step 2: Write the coefficients in the
dividend's place and write the zero of
the linear factor in the divisor's place.
Step 3: Bring the first coefficient
down.
Step 4: Multiply it with the divisor and
write it below the next coefficient.
Step 5: Add them and write the value
below.
Step 6: Repeat the previous 2 steps
until you reach the last term.
Step 7: Separate the last term thus
obtained which is the remainder.
Step 8: Now group the coefficients
with the variables to get the quotient.
Divide 2𝑥2 − 𝑥 − 24 𝑏𝑦 (𝑥 − 3).
2 − 1 − 24
Since the leading term is 𝟐𝒙𝟐, we shall subtract the
exponent to 1 (2 – 1 = 1) which will result to 1. So,
the exponent of the leading term is 1.
3
2
6
5
15
−9
Therefore, 2𝑥 + 5 −
9
𝑥−3
.

15. DIVISIONOFPOLYNOMIALS USING
SYNTHETIC DIVISION
Step 1: Check whether the
polynomial is in the standard form.
Step 2: Write the coefficients in the
dividend's place and write the zero of
the linear factor in the divisor's place.
Step 3: Bring the first coefficient
down.
Step 4: Multiply it with the divisor and
write it below the next coefficient.
Step 5: Add them and write the value
below.
Step 6: Repeat the previous 2 steps
until you reach the last term.
Step 7: Separate the last term thus
obtained which is the remainder.
Step 8: Now group the coefficients
with the variables to get the quotient.
Divide 𝑥3 + 2𝑥2 − 3𝑥 + 1 𝑏𝑦 (𝑥 − 2).
1 2 − 3 1
To know the zero of the divisor or linear factor, equate 𝒙 − 𝟐.
𝑥 − 2
𝑥 − 2 = 0
𝑥 − 2 + 2 = 0 + 2 Addition Property of Equality
𝒙 = 𝟐
2
1
2
4
8
5
10
11
𝒙𝟐
+ 𝟒𝒙 + 𝟓 +
𝟏𝟏
𝒙 − 𝟐

16. Divide 2𝑥3
+ 5𝑥2
+ 9 by (𝑥 + 3).
𝑥 + 3
𝑥 + 3 = 0
𝑥 + 3 − 3 = 0 − 3
𝑥 = −3
2 5 0 9
−3
2
−6
−1
3
3
−9
0
2𝑥2
− 𝑥 + 3

17. SolvethefollowingusingLongDivision
andSyntheticDivision.
1. (5𝑥4 − 𝑥3 + 𝑥 − 2) ÷ (𝑥 + 7)
2. (𝑥4
+ 8𝑥) ÷ (𝑥 + 2).

18. Example: Find the remainder when 𝑷 𝒙 = 𝟑𝒙𝟐
− 𝟏𝟒𝒙 − 𝟏𝟐 is
divided by 𝒙 − 𝟐.
REMAINDER THEOREM
If a polynomial 𝑷(𝒙) is divided by (𝒙 − 𝒓), where 𝒓 is a constant,
then the remainder is 𝑷(𝒓).
𝑃 𝑥 = 3𝑥2 − 14𝑥 − 12
𝑃 2 = 3(2)2−14 2 − 12
𝑃 2 = 3 4 − 28 − 12
𝑃 2 = 12 − 28 − 12
𝑃 2 = −28
To equate the constant,
𝑥 − 2
𝑥 − 2 = 0
𝑥 − 2 + 2 = 0 + 2
𝑥 = 2 ANSWER: Therefore, the remainder is –28.

19. Find the remainder when 𝑷 𝒙 = 𝟑𝒙𝟐 + 𝟕𝒙 − 𝟐𝟎 is divided by 𝒙 + 𝟒.
𝑃 𝑥 = 3𝑥2
+ 7𝑥 − 20
𝑃 −4 = 3(−4)2+7 −4 − 20
𝑃 −4 = 3 16 − 28 − 20
𝑃 −4 = 48 − 28 − 20
𝑃 −4 = 0
To equate the constant,
𝑥 + 4
𝑥 + 4 = 0
𝑥 + 4 − 4 = 0 − 4
𝑥 = −4
ANSWER: Therefore, the remainder is 0.

20. Example: Determine whether 𝒙 − 𝟑 is a factor of the polynomial
𝑷 𝒙 = 𝟐𝒙𝟑
− 𝟓𝒙𝟐
− 𝒙 − 𝟔.
FACTOR THEOREM
For a polynomial 𝑷(𝒙), if 𝑷 𝒓 = 𝟎 for a constant 𝒓, then the
polynomial 𝒙 − 𝒓 is a factor of 𝑷 𝒙 .
𝑃 𝑥 = 2𝑥3 − 5𝑥2 − 𝑥 − 6
𝑃 3 = 2(3)3 − 5 3 2 − 3 − 6
𝑃 3 = 2 27 − 5 9 − 3 − 6
𝑃 3 = 54 − 45 − 3 − 6
𝑃 3 = 0
To equate the constant,
𝑥 − 3
𝑥 − 3 = 0
𝑥 − 3 + 3 = 0 + 3
𝑥 = 3 ANSWER: Therefore, 𝑥 − 3 is a FACTOR of
2𝑥3 − 5𝑥2 − 𝑥 − 6.

21. Determine whether 𝒙 − 𝟑 is a factor of the polynomial 𝑷 𝒙 = 𝟑𝒙𝟑 + 𝒙𝟐 − 𝟐𝒙 − 𝟏𝟓.
To equate the constant,
𝑥 − 3
𝑥 − 3 = 0
𝑥 − 3 + 3 = 0 + 3
𝑥 = 3
𝑃 𝑥 = 3𝑥3 + 𝑥2 − 2𝑥 − 15
𝑃 3 = 3(3)3 + 3 2 − 2 3 − 15
𝑃 3 = 3 27 + 9 − 6 − 15
𝑃 3 = 81 + 9 − 6 − 15
𝑃 3 = 69
ANSWER: Therefore, 𝑥 − 3 is a FACTOR of
𝟑𝒙𝟑
+ 𝒙𝟐
− 𝟐𝒙 − 𝟏𝟓.