Description Identity Example

Sec 3.2 – Polynomial Functions
Factoring Using Polynomial Identities
Name:
Common Polynomial Identities:
Description
Identity
Example
Difference of Two Squares
π‘Ž2 βˆ’ 𝑏 2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)
9π‘₯ 2 βˆ’ 4𝑦 2 = (3π‘₯ + 2𝑦)(3π‘₯ βˆ’ 2𝑦)
Sum of Two Squares
π‘Ž2 + 𝑏 2 = (π‘Ž + 𝑏𝑖)(π‘Ž βˆ’ 𝑏𝑖)
16π‘š2 + 9 = (4π‘š + 3𝑖)(4π‘š βˆ’ 3𝑖)
Perfect Square Trinomial
π‘Ž2 + 2π‘Žπ‘ + 𝑏 2 = (π‘Ž + 𝑏)2
9π‘Ž2 + 24π‘Ž + 16 = (3π‘Ž + 4)2
Perfect Square Trinomial
π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏 2 = (π‘Ž βˆ’ 𝑏)2
25𝑝2 βˆ’ 30π‘π‘ž + 9π‘ž 2 = (5𝑝 + 3π‘ž)2
Binomial Cubed
π‘Ž3 + 3π‘Ž2 𝑏 + 3π‘Žπ‘ 2 + 𝑏 3 = (π‘Ž + 𝑏)3
π‘₯ 3 + 6π‘₯ 2 + 12π‘₯ + 8 = (π‘₯ + 2)3
Binomial Cubed
π‘Ž3 βˆ’ 3π‘Ž2 𝑏 + 3π‘Žπ‘ 2 βˆ’ 𝑏 3 = (π‘Ž βˆ’ 𝑏)3
𝑏 3 βˆ’ 9𝑏 2 + 27𝑏 βˆ’ 27 = (𝑏 βˆ’ 3)3
Difference of Two Cubes
π‘Ž3 βˆ’ 𝑏 3 = (π‘Ž βˆ’ 𝑏)(π‘Ž2 + π‘Žπ‘ + 𝑏 2 )
8𝑀 3 βˆ’ 27 = (2𝑀 βˆ’ 3)(4𝑀 2 + 6𝑀 + 9)
Sum of Two Cubes
π‘Ž3 + 𝑏 3 = (π‘Ž + 𝑏)(π‘Ž2 βˆ’ π‘Žπ‘ + 𝑏 2 )
64𝑦 3 + 1 = (4𝑦 + 1)(16𝑦 2 βˆ’ 4𝑦 + 1)
1. Factor the following using a Difference or Sum of Two Squares.
a.
4π‘Ž2 βˆ’ 25𝑏 2
d. π‘₯ 2 + 36
b. (5π‘š3 )2 βˆ’ (6𝑛)2
c. π‘Ž2 𝑏 8 βˆ’ 9𝑝6 π‘ž 2
e. 18𝑝2 βˆ’ 98π‘ž 2
f. 16𝑀 2 + 7
M. Winking
Unit 3-2 page 45
2. Factor the following using a Difference or Sum of Two Cubes.
π‘Ž3 βˆ’ 64
a.
c. 16π‘š9 βˆ’ 250
b. 27π‘₯ 3 + 8𝑦 6
d. (7𝑝2 )3 + (2π‘ž 4 )3
3. Verify the following polynomial identity from each side.
a.
(π‘Ž2 + 𝑏 2 )2 = (π‘Ž2 βˆ’ 𝑏 2 )2 + (2π‘Žπ‘)2
b. (π‘Ž2 + 𝑏 2 )2 = (π‘Ž2 βˆ’ 𝑏 2 )2 + (2π‘Žπ‘)2
c. This specific identity is commonly used to find sets of Pythagorean triples.
i.
Find the Pythagorean triple that would be created by using a = 3 and b = 2.
ii.
Find the Pythagorean triple that would be created by using a = 5 and b = 2.
M. Winking
Unit 3-2 page 46
4. Verify the following polynomial identity.
π‘₯ 2 + 𝑦 2 + 𝑧 2 + 2(π‘₯𝑦 + π‘₯𝑧 + 𝑦𝑧) = (π‘₯ + 𝑦 + 𝑧)2
M. Winking
Unit 3-2 page 47