1. 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.
2.2.1: Polynomial Identities

2. 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.
2.2.1: Polynomial Identities

3. 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)
2.2.1: Polynomial Identities

4. 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)
2.2.1: Polynomial Identities

5. 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)
2.2.1: Polynomial Identities

6. 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.
2.2.1: Polynomial Identities

7. 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
2.2.1: Polynomial Identities

8. Guided Practice Example 1
Use a polynomial identity to expand the expression (x – 14)2.
2.2.1: Polynomial Identities

9. Guided Practice: Example 1, continued
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.
2.2.1: Polynomial Identities

10. Guided Practice: Example 1, continued
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.
2.2.1: Polynomial Identities

11. Guided Practice: Example 1, continued
The rewritten identity is (x – 14)2 = x2 – 2(x)(14)
(x – 14)2
Original expression
(a – b)2 = a2 – 2ab + b2
Square of Differences Identity
[(x) – (14)]2 = (x)2 – 2(x)(14) + (14)2
Substitute x for a and 14 for b in the Square of Differences Identity.
2.2.1: Polynomial Identities

12. Guided Practice: Example 1, continued
Simplify the expression by finding any products and evaluating any exponents.
The first term, x2, cannot be simplified further, but the other terms can.
2.2.1: Polynomial Identities

13. ✔ Guided Practice: Example 1, continued
When expanded, the expression (x – 14)2 can be written as x2 – 28x
(x – 14)2 = x2 – 2(x)(14) + 142
Equation from step 2
= x2 – 28x + 142
Simplify 2(x)(14).
= x2 – 28x + 196
Square 14. ✔
2.2.1: Polynomial Identities

14. Guided Practice: Example 1, continued
2.2.1: Polynomial Identities

15. Guided Practice Example 2
Use a polynomial identity to factor the expression x
2.2.1: Polynomial Identities

16. Guided Practice: Example 2, continued
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 can be rewritten as x
2.2.1: Polynomial Identities

17. Guided Practice: Example 2, continued
The identity a3 + b3 = (a + b)(a2 – ab + b2) is in the same form as x
Therefore, the expression x , rewritten as
x , is in the same form as the left side of the Sum of Two Cubes Identity: a3 + b3 = (a + b)(a2 – ab + b2).
2.2.1: Polynomial Identities

18. Guided Practice: Example 2, continued
Replace a and b in the identity with the terms in the given expression.
For the rewritten expression x , let x = a and 5 = b in the Sum of Two Cubes Identity.
2.2.1: Polynomial Identities

19. Guided Practice: Example 2, continued
The rewritten identity is x = (x + 5)[x2 – (x)(5) + 52].
x3 + 53
Rewritten expression
a3 + b3 = (a + b)(a2 – ab + b2)
Sum of Two Cubes Identity
(x)3 + (5)3 = [(x) + (5)][(x)2 – (x)(5) + (5)2]
Substitute x for a and 5 for b in the Sum of Two Cubes Identity.
2.2.1: Polynomial Identities

20. Guided Practice: Example 2, continued
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.
2.2.1: Polynomial Identities

21. ✔ Guided Practice: Example 2, continued
When factored, the expression x , which is equivalent to x3 + 53, can be rewritten as
(x + 5)(x2 + 5x + 25).
x = (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. ✔
2.2.1: Polynomial Identities

22. Guided Practice: Example 2, continued
2.2.1: Polynomial Identities