Introduction
Polynomials are often added, subtracted, and even
multiplied. Doing so results in certain sums, differences,
or products of polynomials appearing more commonly.
Becoming familiar with certain polynomial identities can
make the processes of expanding and factoring
polynomials easier.
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2A.2.1: Polynomial Identities
Key Concepts
• Identities can be used to expand or factor polynomial
expressions.
• A polynomial identity is a true equation that is often
generalized so it can apply to more than one example.
• The tables that follow shows the most common
polynomial identities, including the steps of how to
work them out.
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2A.2.1: Polynomial Identities
Key Concepts, continued
Common Polynomial Identities
Square of Sums Identity
Formula
Steps
With two variables:
(a + b)2 = a2 + 2ab + b2
(a + b)2
= (a + b)(a + b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
With three variables:
(a + b + c)2 = a2 + b2 + c2 +
2ab + 2bc + 2ac
(a + b + c)2
= (a + b + c)(a + b + c)
= a2 + ab + ac + ab + b2 + bc + ac + bc + c2
= a2 + b2 + c2 + 2ab + 2bc + 2ac
(continued)
2A.2.1: Polynomial Identities
3
Key Concepts, continued
Square of Differences Identity
Formula
Steps
(a – b)2 = a2 – 2ab + b2
(a – b)2
= (a – b)(a – b)
= a2 – ab – ab + b2
= a2 – 2ab + b2
Difference of Two Squares Identity
Formula
Steps
a2 – b2 = (a + b)(a – b)
a2 – b2
= a2 + ab – ab + b2
= (a + b)(a – b)
(continued)
2A.2.1: Polynomial Identities
4
Key Concepts, continued
Sum of Two Cubes Identity
Formula
Steps
a3 + b3 = (a + b)(a2 – ab + b2)
a3 + b3
= a3 – a2b + ab2 + a2b – ab2 + b3
= (a + b)(a2 – ab + b2)
Difference of Two Cubes Identity
Formula
Steps
a3 – b3 = (a – b)(a2 + ab + b2)
a3 – b3
= a3 + a2b + ab2 – a2b – ab2 – b3
= (a – b)(a2 + ab + b2)
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2A.2.1: Polynomial Identities
Key Concepts, continued
• Note that the square of sums, (a + b)2, is different
from the sum of squares, a2 + b2.
• Polynomial identities are true for any values of the
given variables.
• When dealing with large numbers, the use of identities
often results in simpler calculations.
• The Common Polynomial Identities are found on page
189 in you Unit 2A workbook.
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2A.2.1: Polynomial Identities
Common Errors/Misconceptions
• confusing identities, such as applying the Square of
Differences Identity instead of the Difference of Two
Squares Identity
• incorrectly applying an identity to find an equivalent
expression
• misplacing a negative sign in an identity, resulting in an
incorrect equivalent expression
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2A.2.1: Polynomial Identities
Guided Practice
Example 1
Use a polynomial identity to expand the expression
(x – 14)2.
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2A.2.1: Polynomial Identities
Guided Practice: Example 1, continued
1. Determine which identity is written in the
same form as the given expression.
The expression (x – 14)2 is written in the same form
as the left side of the Square of Differences Identity:
(a – b)2 = a2 – 2ab + b2.
Therefore, we can substitute the values from the
expression (x – 14)2 into the Square of Differences
Identity.
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2A.2.1: Polynomial Identities
Guided Practice: Example 1, continued
2. Replace a and b in the identity with the
terms in the given expression.
For the given expression (x – 14)2, let x = a and
14 = b in the Square of Differences Identity.
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2A.2.1: Polynomial Identities
Guided Practice: Example 1, continued
(x – 14)2
(a –
b)2
=
Original expression
a2
– 2ab +
b2
[(x) – (14)]2 = (x)2 – 2(x)(14) + (14)2
Square of
Differences Identity
Substitute x for a
and 14 for b in
the Square of
Differences Identity.
The rewritten identity is (x – 14)2 = x2 – 2(x)(14) + 142.
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2A.2.1: Polynomial Identities
Guided Practice: Example 1, continued
3. Simplify the expression by finding any
products and evaluating any exponents.
The first term, x2, cannot be simplified further, but the
other terms can.
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2A.2.1: Polynomial Identities
Guided Practice: Example 1, continued
(x – 14)2 = x2 – 2(x)(14) + 142
Equation from step 2
= x2 – 28x + 142
Simplify 2(x)(14).
= x2 – 28x + 196
Square 14.
When expanded, the expression (x – 14)2
can be written as x2 – 28x + 196.
✔
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2A.2.1: Polynomial Identities
Guided Practice: Example 1, continued
14
2A.2.1: Polynomial Identities
Guided Practice
Example 2
Use a polynomial identity to factor the expression x 3 + 125.
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2A.2.1: Polynomial Identities
Guided Practice: Example 2, continued
1. Determine which identity is written in the
same form as the given expression.
The first term in the expression, x 3, is a cube.
Determine whether the second term in the
expression, 125, is also a cube.
125 = 5 • 25 = 5 • 5 • 5 = 53
Since 125 can be rewritten as a cube, the expression
x 3 + 125 can be rewritten as x 3 + 53.
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2A.2.1: Polynomial Identities
Guided Practice: Example 2, continued
The identity a3 + b3 = (a + b)(a2 – ab + b2) is in the
same form as x 3 + 53.
Therefore, the expression x3 + 125, rewritten as
x 3 + 53, is in the same form as the left side of the Sum
of Two Cubes Identity: a3 + b3 = (a + b)(a2 – ab + b2).
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2A.2.1: Polynomial Identities
Guided Practice: Example 2, continued
2. Replace a and b in the identity with the
terms in the given expression.
For the rewritten expression x 3 + 53, let x = a and
5 = b in the Sum of Two Cubes Identity.
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2A.2.1: Polynomial Identities
Guided Practice: Example 2, continued
Rewritten
expression
x 3 + 53
a3
+
b3
= (a +
b)(a2
– ab +
b 2)
Sum of Two
Cubes Identity
Substitute x for
a and 5 for b in
(x)3 + (5)3 = [(x) + (5)][(x)2 – (x)(5) + (5)2]
the Sum of Two
Cubes Identity.
The rewritten identity is x3 + 53 = (x + 5)[x2 – (x)(5) + 52].
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2A.2.1: Polynomial Identities
Guided Practice: Example 2, continued
3. Simplify the expression by finding any
products and evaluating any exponents.
The first factor, x + 5, cannot be simplified further.
The second factor, x2 – (x)(5) + 52, has three terms:
the first term, x2, cannot be simplified; the other terms,
(x)(5) and 52, can be simplified.
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2A.2.1: Polynomial Identities
Guided Practice: Example 2, continued
x3 + 53 = (x + 5)[x2 – (x)(5) + 52]
Equation from
the previous step
= (x + 5)(x2 – 5x + 52)
Simplify (x)(5).
= (x + 5)(x2 + 5x + 25)
Square 5.
When factored, the expression x3 + 125, which is
equivalent to x3 + 53, can be rewritten as
(x + 5)(x2 + 5x + 25).
✔
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2A.2.1: Polynomial Identities
Guided Practice: Example 2, continued
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2A.2.1: Polynomial Identities
Example 3
Use a polynomial identity to expand the
expression 3𝑥 2 − 2𝑥 + 8 2 .
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2A.2.1: Polynomial Identities
Example 3 continued…
Use a polynomial identity to expand the expression 3𝑥 2 − 2𝑥 + 8 2 .
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2A.2.1: Polynomial Identities
Example 3 continued…
Use a polynomial identity to expand the expression 3𝑥 2 − 2𝑥 + 8 2 .
25
2A.2.1: Polynomial Identities
Example 3 continued…
Use a polynomial identity to expand the expression 3𝑥 2 − 2𝑥 + 8 2 .
26
2A.2.1: Polynomial Identities
Example 3 continued…
Use a polynomial identity to expand the expression 3𝑥 2 − 2𝑥 + 8 2 .
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2A.2.1: Polynomial Identities
Example 3 continued…
Use a polynomial identity to expand the expression 3𝑥 2 − 2𝑥 + 8 2 .
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2A.2.1: Polynomial Identities
Example 4
Show how 362 can be evaluated using
polynomial identities.
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2A.2.1: Polynomial Identities
Example 4 Continued…
Show how 362 can be evaluated using polynomial identities.
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2A.2.1: Polynomial Identities
Example 4 Continued…
Show how 362 can be evaluated using polynomial identities.
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2A.2.1: Polynomial Identities
Example 4 Continued…
Show how 362 can be evaluated using polynomial identities.
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2A.2.1: Polynomial Identities
Practice problems…
Use polynomial identities to expand or factor each expression.
#3
𝟏𝟎𝟎𝒙𝟐 − 𝟒
1. Determine which identity is written in the same form as the
given expression.
𝟏𝟎𝟐 𝒙𝟐 − 𝟐𝟐
Difference of two square identities.
2. Replace a and b in the identity with the terms in the given
expression:
𝟏𝟎𝟐 𝒙𝟐 − 𝟐𝟐 = 𝟏𝟎𝒙 + 𝟐 𝟏𝟎𝒙 − 𝟐 = 𝟐 𝟓𝒙 + 𝟏 𝟓𝒙 − 𝟏
33
2A.2.1: Polynomial Identities
Practice Problems…
Use polynomial identities to expand or factor each expression.
#7 𝒙𝟐 + 𝒙 + 𝟑
𝟐
1. Polynomial Identity: Square of Sum with three variables:
2. Replace a and b in the identity with the terms in the given
expression.
= 𝒙𝟐
𝟐
3. Simplify.
+ 𝒙
𝟐
+ 𝟑
𝟐
+ 𝟐 𝒙𝟐 𝒙 + 𝟐 𝒙𝟐 𝟑 + 𝟐 𝒙 𝟑
= 𝒙𝟒 + 𝒙𝟐 + 𝟗 + 𝟐𝒙𝟑 + 𝟔𝒙𝟐 + 𝟔𝒙
= 𝒙𝟒 + 𝟐𝒙𝟑 + 𝟕𝒙𝟐 + 𝟔𝒙 + 𝟗
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2A.2.1: Polynomial Identities
Practice Problems…
Find the area of a square with the given side length, without using a
calculator. The area of a square is the square of the side length, or
𝑎𝑟𝑒𝑎 = 𝑠𝑖𝑑𝑒 2 .
470m
9. side length = 470 meters
𝑎 + 𝑏 2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2
0 + 470 2 = 02 + 2 0 470 + 4702
470 𝑚 ∗ 470 𝑚 = 220,900 𝑚2
470m
Another way:
400 + 70 2 = 4002 + 2 400 70 + 702
= 160000 + 2 28000 + 4900
= 160000 + 56000 + 4900
= 220,900 𝑚2
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2A.2.1: Polynomial Identities