Division of Polynomials
Topic 1: Polynomials
Methods used for dividing polynomials:
1.) Quotient Rule
Addition and Subtraction of Polynomials
When adding or subtracting polynomials, add or subtract all like terms
accordingly
𝑚
𝑚
𝑎
𝑛
𝑚−𝑛
𝑜𝑟 𝑎 / 𝑎 = 𝑎
𝑛
𝑎
Ex.
3
Example of Addition:
−49𝑥
2
7𝑥
1.)
2.)
2
2
(4𝑥 + 8𝑥 + 2) + (− 3𝑥
2
+ 4𝑥 − 7)
3
2
2
42𝑥
7
−
2
7𝑥
↓
↓
7𝑥 +
6𝑥
84𝑥
2
7𝑥
↓
3
5
+ 12𝑥
3
(4𝑥 + 5 − 𝑥 + 2𝑥) + (6 − 5𝑥 + 2𝑥 )
2
5
+
2
3
4𝑥 - 3𝑥 = 𝑥
Long Division Method:
If a there is a missing term add a 0
3
𝑥 − 2𝑥
2
8𝑥 + 4𝑥 = 12𝑥
2 - 7 = -5
2
4𝑥 − 5𝑥
2𝑥 = 2𝑥
6 + 5 = 11
2
3
= 𝑥 + 12𝑥 - 5
2
= 𝑥 - 𝑥 + 2𝑥 + 11
Example of Subtraction:
1.)
2.)
2
(4𝑥 − 2𝑥 + 3𝑥 − 7) - (3𝑥 − 5 + 𝑥 )
↓
3
2
3
(5𝑥 + 3𝑥 − 6) - (7𝑥 − 8)
↓
2
2
2
2
(5𝑥 + 3𝑥 − 6) + (− 7𝑥 + 8)
(4𝑥 − 2𝑥 + 3𝑥 − 7) + (− 3𝑥 + 5 − 𝑥 )
↓
By law of subtraction, reciprocate signs of second term
2
2
3
5𝑥 = 5𝑥
3
4𝑥 = 4𝑥
2
2
2
3𝑥 − 7𝑥 = 4𝑥
− 𝑥 + 3𝑥 = 2𝑥
8− 6 = 2
− 2𝑥 − 3𝑥 = − 5𝑥
−7+5= −2
2
= 4𝑥 + 2𝑥 − 5𝑥 − 2
3
= 5𝑥 − 4𝑥 + 2
2
Methods used for multiplying polynomials:
1.) Product rule
𝑛
𝑚+𝑛
3
4
7
𝑎 . 𝑎 =𝑎
𝑎 .𝑎 = 𝑎
Ex.
2
2𝑎 (𝑎 + 3) = 2𝑎 + 6𝑎
3.) Foil Method (First, Outside, Inside, Last)
(𝑎 + 𝑏) (𝑐 + 𝑑) = 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑
Ex.
2
(3𝑥 + 2) (4𝑥 − 5) = 12𝑥 − 15𝑥 + 8𝑥 − 10
Example of Multiplying:
1.)
2.)
2
(− 2𝑥 − 4𝑥 + 1) (𝑥 − 3𝑥 − 2)
Distributive Property:
3
− 2𝑥 =
2
− 4𝑥 =
1
5
3
(3𝑥 + 2) (4𝑥 − 5)
Foil Method:
4
3
− 2𝑥 + 6𝑥 + 4𝑥
2
− 4𝑥 + 12𝑥 + 8𝑥
2
= 𝑥 − 3𝑥 − 2
2
(3𝑥) 4𝑥 = 12𝑥
3𝑥 (− 5) =
− 15𝑥
(2) 4𝑥 = 8𝑥
(2) − 5 = − 10
Then add all like terms:
5
4
Synthetic Division
Shortcut for long division as long as the divisor is “𝑥 − 𝑛”
Synthetic division cannot be used when the divisor has powers or a
negative exponent
Polynomials must be in descending order and ensure that all powers are
represented
Example:
2.) Distributive Property
𝑎 (𝑏 + 𝑐) = 𝑎𝑏 + 𝑎𝑐
3
Topic 2: Dividing Polynomials using Synthetic Division
Steps on how to do Synthetic Division
1.) Write the coefficient of the dividend and use the zero of the linear factor in
the divisor’s place
2.) Bring the coefficient down and multiply with the divisor
3.) Write the product below the second coefficient
4.) Repeat with the last coefficient
Multiplication of polynomials
𝑚
____________________________________________________________________
2
= − 2𝑥 − 6𝑥 + 13𝑥 + 5𝑥 − 2
2
= 12𝑥 − 15𝑥 + 8𝑥 − 10
Topic 4: Evaluating Polynomial Function
Topic 6: Zero Polynomial Function
Polynomial Function
A polynomial function of degree “n” is defined by
𝑛
𝑛−1
𝑓(𝑥) = 𝑎𝑛𝑥 + 𝑎𝑛−1𝑥
-
. . . + 𝑎1𝑥 + 𝑎0
For real numbers an 𝑎𝑛−1 . . . , 𝑎1& 𝑎0. Where 𝑎𝑛 ≠ 0 and “n” is a whole
Zero Polynomial Function
A polynomial whose value is zero
It is a constant polynomial with a constant function of value 0
If the factor’s quotient is equal to 0, it is considered to be one of the zeros
number
Example:
1.)
Example:
1.)
2.)
3
2
3
2
3
𝑓(𝑥) = 4𝑥 − 5𝑥 + 7, 𝑓𝑖𝑛𝑑 𝑓(2)
𝑓(𝑥) = 4𝑥 − 5𝑥 + 7, 𝑓𝑖𝑛𝑑 𝑓(− 3)
Note: Follow PEMDAS rule
3
List the possible values of
𝑝 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡): ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24
List the possible values of 𝑞 (𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒): ± 1
2
𝑓(2) = 4(2) − 5(2) + 7
3
2
𝑥 + 9𝑥 − 26𝑥 − 24
2
𝑓(2) = 4(− 3) − 5(− 3) + 7
Possible rational roots
𝑓(2) = 4(8) − 5(4) + 7
𝑓(2) = 32 − 20 + 7
𝑓(2) = 4(− 27) − 5(9) + 7
𝑓(2) =− 108 − 45 + 7
𝑓(2) = 19
𝑓(− 3) =
𝑝
𝑞
: ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24
a.
1
− 146
−9
1
−8
1
____________________________________________________________________
26 − 24
−8
18
18 − 6
∟ 1
→ 1 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑧𝑒𝑟𝑜
b.
Topic 5: Remainder & Factor Theorem
1
Theorem
A proven or probable rule or concept in Mathematics, established by
relying on previously accepted truths
Acts as a foundational element in Mathematics, helping us understand the
connections between various mathematical concepts and how they
interrelate
−9
26 − 24 ∟ 2
2 − 14 − 24
→ 2 𝑖𝑠 𝑎 𝑧𝑒𝑟𝑜
−7
12
0
↓ 𝐷𝑒𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
1
2
𝑥 − 7𝑥 + 12
To find the other factors, find the binomial of the depressed equation:
Find the factors that equals to the sum and product of the equation:
− 4 (− 3) = 12
− 4 + (− 3) = 7
Remainder Theorem
If a polynomial 𝑓(𝑥), is divided by (𝑥 − 𝑎), the remainder is the constant
𝑓(𝑎), and:
𝑓(𝑥) = 𝑞(𝑥) • (𝑥 − 𝑎) + 𝑓(𝑎)
Where 𝑞(𝑥) is a polynomial with degree one less than the degree of 𝑓(𝑥)
Zeroes: 2, − 4, − 3
Roots: (𝑥 − 2), (𝑥 + 4), (𝑥 + 3)
Example:
2.)
Now, in Synthetic Division
2
3
2
𝑥 + 3𝑥 − 4𝑥 − 12
2
𝑓(𝑥) = 5𝑥 − 4𝑥 + 3, 𝑓(3)
𝑓(𝑥) = 5𝑥 − 4𝑥 + 3, 𝑓(3)
2
𝑓(3) = 5(3) − 4(3) + 3
𝑓(3) = 5(9) − 4(3) + 3
𝑓(3) = 45 − 12 + 3
← 𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
5
−4
15
11
5
3 ∟ 3
33
36
List the possible values of 𝑝 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡): ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12
List the possible values of 𝑞 (𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒): ± 1
Possible rational roots
𝑝
𝑞
: ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12
a.
1
𝑓(3) = 36
3
1
4
𝑓(3) = 36
1
Note: The value obtained when evaluating the function at 𝑓(3) and the value of the
remainder when dividing the polynomial by (𝑥 − 3) are the same
∟ 1
→ 1 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑧𝑒𝑟𝑜
b.
1
Factor Theorem
The binomial (𝑥 − 𝑎) is a factor of the polynomial 𝑓(𝑥) if and only if
𝑓(𝑎) = 0
− 4 − 12
4
0
0 − 12
−1
3
− 4 − 12
−1 −2
6
2 − 6 −6
∟ −1
→ − 1 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑧𝑒𝑟𝑜
c.
3
2
1
Example: 𝑝(𝑎) = 𝑥 + 4𝑥 − 15𝑥 − 18, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑠
3
2
4
1
𝑝(5)
𝑝(5)
− 4 − 12 ∟ 2
10
12
→ 2 𝑖𝑠 𝑎 𝑧𝑒𝑟𝑜
6
0
↓ 𝐷𝑒𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
2
3
2
𝑝(5) = 𝑥 + 4𝑥 − 15𝑥 − 18
3
3
𝑥 + 5𝑥 + 6
2
𝑝(5) = 𝑥 + 4𝑥 − 15𝑥 − 18
2
1
4
− 15 − 18 ∟ 3
5
45
150
1
9
30 132
𝑝(5) = 132, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑖𝑡 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑝(5) = 132, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑖𝑡 𝑖𝑠 𝑁𝑂𝑇 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟
(5) + 4(5) − 15(5) − 18
125 + 100 − 75 − 18
𝑝(3)
𝑝(− 6)
3
2
𝑝(3) = 𝑥 + 4𝑥 − 15𝑥 − 18
3
2
(3) + 4(3) − 15(3) − 18
27 + 36 − 45 − 18
𝑝(3) = 0, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑖𝑡 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟
3
2
𝑝(− 6) = 𝑥 + 4𝑥 − 15𝑥 − 18
1
∟ −6
4
− 15 − 18
−6
12
18
1
9
−3
0
𝑝(− 6) = 0, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑖𝑡 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟
2 ( 3) = 6
2+3 = 5
Zeroes: 2, − 2, − 3
Roots: (𝑥 − 2), (𝑥 + 2), (𝑥 + 3)