DIVIDING OF POLYNOMIALS

1. Division of
Polynomials,
Remainder and
Factor Theorems

2. OBJECTIVES:
1. Divides polynomials using Long Division.
2. Divides polynomials using Synthetic Division.
3. Finds the remainder when the given polynomial
is divided by (x – c) using Remainder Theorem.
4. Uses the Factor Theorem to determine whether or
not the given binomial (x – c) is a factor of the given
polynomial.
5. Applies Factor Theorem and Remainder Theorem
in solving problems. (ie. find the value of k...)

3. Division of Polynomials
To divide polynomials, we use long division, as follows.
DIVISION ALGORITHM:
If P(x) and D(x) are polynomials, with D(x) 0, then there exists
a unique polynomial Q(x) and R(x) , where R(x) is either 0 or of
degree less than the degree of D(x), such that:
= Q(x) + or P(x) = D(x) Q(x) + R(x)
Dividend = Divisor Quotient + Remainder
The polynomials P(x) and D(x) are called the dividend and
divisor, respectively; Q(x) is the quotient, and R(x) is the
remainder.

4. A. LONG DIVISION
Example 1. Divide (2x3
– 5x2
– 8x + 15) by (x – 3).
Quotient:
2x2
+ x – 5

5. Example 2. Divide (3x4
– 12x + 5) by (x + 1).
Solution:
Quotient:
3x3
– 3x2
+ 3x – 15 +

6. Example 3. Divide (6x3
+ 11x2
– 31x + 1) by (3x – 2).
Solution:
Quotient:
2x2
+ 5x – 7 –

7. Example 4. Divide (3x4
+ x3
– 17x2
+ 19x – 6 ) by (x2
–2x + 1).
Solution:
Quotient:
3x2
+ 7x – 6

8. Example 5. Divide (2x3
– 3x2
+ 4x + 5 ) by (x + 2).
Solution:
Quotient:
2x2
-7x + 18 -

9. Example 6. Divide (8x4
+ 6x2
– 3x + 1 ) by (2x2
– x + 2).
Solution:
Quotient:
4x2
+ 2x +

10. Example 7. Divide (– 3x3
+ 7x2
– 11x – 1 ) by (3x – 1).
Solution:
Quotient:
– x2
+ 2x – 3 –

11. B. SYNTHETIC DIVISION
Example 1. Divide (2x3
– 5x2
– 8x + 15) by (x – 3).
Solution:
3 2 -5 – 8 15
6 3 -15
2 1 -5 0 Remainder
Quotient: 2x2
+ x – 5

12. Example 2. Divide (3x4
– 12x + 5) by (x + 1).
Solution:
-1 3 0 0 -12 5
-3 3 - 3 15
3 -3 3 -15 20 Remainder
Quotient: 3x3
– 3x2
+3x – 15 +

13. Example 3. Divide (6x3
+ 11x2
– 31x + 1) by (3x – 2)
Solution:
6 11 -31 1
2 4 10 -14
6 15 -21 -13 Remainder
3 2 5 -7
Quotient: 2x2
+ 5x – 7 –

14. Example 4. Divide (2x3
– 3x2
+ 4x + 5) by (x + 2)
Solution:
- 2 -3 4 5
-4 14 -36
2 -7 18 -31 Remainder
Quotient: 2x2
– 7x + 18 –

15. Example 5. Divide (– 3x3
+ 7x2
– 11x – 1) by (3x – 1)
Solution:
– 3 7 –11 – 1
-1 2 – 3
–3 6 –9 – 4 Remainder
3 -1 2 -3
Quotient: – x2
+ 2x – 3 –

16. Example 6. Divide (3x4
+ x3
– 17x2
+ 19x – 6) by (x2
– 2x + 1)
Solution:
3 1 -17 19 -6
6 -3 -7 6
14 -12
3 7 -6 0 0
Quotient: 3x2
+ 7x – 6

17. Example 7. Divide (8x4
+ 6x2
– 3x + 1) by (2x2
– x + 2)
Solution:
8 0 6 –3 1
4 -8 -4
2
8 4 0 -7 1
2 4 2
Quotient: 4x2
+ 2x +

18. The Remainder Theorem:
If the polynomial P (x) is divided by (x – c), the remainder R is a constant and is
equal to P(c). Hence, R = P(c)
Thus, there are two ways to find the remainder when P(x) is divided
by (x – c), that is:
1. Use synthetic division, or
2. Calculate P(c).
Similarly, there are two ways to find the value of P(c):
1. Substitute c in the polynomial expression P(x), or
2. Use synthetic division.
The computation 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).

19. Examples: Find the remainder when the polynomial P(x) is divided by (x – c):
1. P(x) = 2 + 5x – 5 is divided by ( x + 2).
Solution 1: By using remainder theorem
P(-2) = 2 + 5(-2) – 5 = 2(-8) – 4(4) – 10 – 5 = – 16 – 16 – 15
P(-2) = – 47
Hence, the remainder is – 47.
Solution 2: By using synthetic division
-2 2 -4 5 -5
-4 16 -42
2 -8 21 -47 Remainder

20. 2. P(x) = 3 + 4 – 3x + 4 is divided by ( x – 1 ).
Solution 1: By using remainder theorem
P(1) = 3 + 4 – 3(1) + 4 = 3(1) – 6(1) + 4(1) – 3(1) + 4
= 3 – 6 + 4 – 3 + 4 = 2
Hence, the remainder is 2.
Solution 2: By using synthetic division
1 3 -6 4 -3 4
3 -3 1 -2
3 -3 1 -2 2 Remainder

21. 3. P(x) = 4 + 4 - 7x + 6 is divided by (x +3).
Solution 1: By using remainder theorem
P(-3) = 4 + 4 – 7(-3) + 6
= 4(-243) – 2(-27) + 4(9) + 21 + 6 = -972 + 54 + 36 + 21 + 6
= -972 + 117 = -855
Hence, the remainder is -855.
Solution 2: By using synthetic division
-3 4 0 -2 4 -7 6
-12 36 -102 294 -861
4 -12 34 -98 287 -855 Remainder

22. 4. P(x) = 5 + 6 - 7 + 6x – 4 is divided by (x +1).
Solution : By using remainder theorem
P(-1) = 5 + 6 – 7 + 6(-1) – 4
= 5(-1) – 3(1) + 6(-1) – 7(1) – 6 – 4
= -5 – 3 – 6 – 7 – 6 – 4 = -37
Hence, the remainder is –37.
5. P(x) = 2 + 3 - 8x + 6 is divided by (2x +3).
Solution : By using remainder theorem
P = + 6
= 2 + 6
= + + + 12 + 6 = + + + 18 = + = =
Hence, the remainder is

23. The Factor Theorem:
The polynomial P(x) has x – r as a factor if and only if P(r) = 0.
Proof:
There are two parts of the proof of the Factor theorem, namely:
1. If x – r is a factor of P(x), then P(r) = 0.
2. If P(r) = 0, then (x –r) is a factor of P(x).
Example 1: Show that (x – 2) is a factor of P(x) = 2 - 2x + 24.
Solution:
P(2) = 2 = 2(8) – 9(4) – 4 + 24 = 16 – 36 + 20
= 36 – 36 = 0
Hence, (x – 2) is a factor of P(x)

24. Example 2: Determine whether (x – 3) is a factor of
P(x) = 4 + 6x – 9 .
Solution:
P(3) = 4 = 4(27) – 3(9) + 18 – 9
= 108 – 27 + 9 = 90
Since the remainder is 90 and not equal to zero,
then (x – 3) is not a factor of P(x).
Example 3: Is (5x + 1) a factor of P(x) = 5 – 29x – 6?
Solution:
P = 5 - 6
= 5 +6 + - 6 = + + - 6
= - 6 = - 6 = 6 – 6 = 0
Clearly, the remainder is zero, then (5x + 1) is a factor of P(x).

25. Example 4: Determine whether the given polynomial P(x) is Divisible
by the given binomial (x – r)
a) P(x) = 2 + 8x – 4 ; (x + 2)
Solution:
P(-2) = 2
= -16 – 28 – 20 = -64
Since the remainder is – 64, then P(x) is not Divisible by (x + 2)
b) P(x) = 2 – 5x + 6 ; (x – 3)
Solution:
P(3) = 2
= 162 – 27 – 126 – 9 = 162 – 162 = 0
Since the remainder is 0, then P(x) is Divisible by (x – 3 )

26. Example 5: Find the remainder when P(x) is divided by the D(x) and determine
if D(x) is a factor of P(x).
P(x) D(x) Remainder
R(x)
Factor or
Not Factor?
1. 3x3
– 17x2
+ 18x + 8 x – 2
2. 2x3
– 3x2
– 4x + 6 x + 2
3. 2x4
- 11x3
+10x2
+ 11x – 12 x – 4
4. 4x4
+ 4x3
-8x2
+ 9x + 6 x + 3

27. Applications of Remainder and Factor Theorems:
Example 6: Find the value of k so that (x + 2) is a factor of:
a) 3 + 3kx + 5
Solution: P(-2) = 0
P(-2) = 3
0 = -24 -8k -6k + 5
14k = -19; k =
b) 2k – 6x + 4
Solution: P(-2) = 0
P(-2) =
0 = 32k + 40 + 12 + 16
-32k = 68; k = -

28. Example 7: Find the value of k so that – 4 is the remainder when
6k - 5x + 4 is divided by x + 2?
Solution: P(-2) = -4
P(-2) = 6k
-4 = -48k + 12k + 14 -4 = -36k + 14 -36k = -18
k =
Example 8: Find the value of k so that x + 2 is a factor
of 2k – 6x + 4.
Solution: P(-2) = 0
P(-2) =
0 = 32k + 40 + 12 + 16
-32k = 68; k = -

29. Example 9: Find the values of A & B so that x + 1 is a factor of
P(x) = 3A + 4x + 5 and when divided by x – 2 the remainder is 6.
Solution:
i) P(-1) = 0 and ii) P(2) = 6
P(-1) = 3A
0 = -3A + 2B + 1 3A – 2B = 1 (Eq. 1)
P(2) = 3A
6 = 24A + 8B + 13 24A + 8B = -7 (Eq.2)
24A + 8B = -7 3A – 2B = 1 12A + 5 = 4
-8 (3A – 2B = 1) 3A– 2 = 1 12A = -1
24A + 8B = -7 3A + = 1 A =
-24A + 16B = -8
24B = -15
B =

30. Example 10: Find the values of A & B so that x – 1 and x + 2 are both factors of
P(x) = 4A + 2x + 6.
Solution:
i) P(1) = 0 and ii) P(-2) = 0
P(1) = 4A
0 = 4A - 3B + 8 4A – 3B = -8 (Eq. 1)
P(-2) = 4A
0 = -32A - 12B + 2 -32A -12B = -2 (Eq.2)
-32A -12B = -2 4A – 3B = -8 8A -11 = -16
8(4A – 3B = -8) 4A– 3 = -8 8A = -16+11 8A = -5
-32A -12B = -2 4A - = -8 A =
32A -24B = -64
-36B = -66
B =

31. 𝑻𝑯𝑨𝑵𝑲𝒀𝑶𝑼 …