HSN.CN.C.8Common CoreMathNumber and QuantityGrades 9-12
HSN.CN.C.8: Extending Polynomial Identities to the Complex Numbers
In plain English: HSN.CN.C.8 is an advanced (+) Common Core number and quantity standard that asks students to extend polynomial identities, such as the difference of squares, to the complex numbers. The official example rewrites x² + 4 as (x + 2i)(x - 2i), so a sum of squares factors into linear factors. It is usually taught in Algebra II or Precalculus.
(+) Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x - 2i).
Common Core State Standards for Mathematics · Domain: The Complex Number System (CN) · Cluster: Use complex numbers in polynomial identities and equations. Also written as HSN-CN.C.8 or N-CN.8 · Official standard
Students see that polynomial identities they already know, such as A² - B² = (A + B)(A - B) and (A + B)² = A² + 2AB + B², remain true when A or B is a complex number. The key idea is that these identities follow from the properties of addition and multiplication, and complex numbers have the same properties. With i² = -1, the difference of squares turns into a factorization of a sum of squares, as in the official example x² + 4 = (x + 2i)(x - 2i).
Students then extend the square of a binomial and the product of conjugate linear factors, factor quadratics such as x² + 2x + 10 into linear factors over the complex numbers, and check each result by multiplying it back out.
Learning Objectives
By the end of this lesson, students will be able to:
Explain why a polynomial identity that holds for real numbers also holds for complex numbers
Rewrite a sum of squares such as x² + 4 as a product of conjugate linear factors, as in the official example
Expand squares of binomials with an imaginary term, such as (x + 3i)², using i² = -1
Factor quadratics with real coefficients into linear factors over the complex numbers by completing the square
Verify a complex factorization by multiplying the factors back out
Prior Knowledge Required
Students should already be comfortable with:
Adding, subtracting and multiplying complex numbers with i² = -1 HSN.CN.A.2
Using the structure of an expression, such as the difference of squares, to rewrite it HSA.SSE.A.2
Proving polynomial identities by rewriting one side HSA.APR.C.4
Completing the square in a quadratic expression HSA.SSE.B.3
Ask students to factor two expressions and then compute two powers of i-multiples. Give four minutes.
Warm-Up Prompt
"Factor x² - 9. Then try to factor x² + 9 over the real numbers. Finally compute (3i)² and (-3i)². How could your last two answers help you factor x² + 9?"
Students factor x² - 9 = (x + 3)(x - 3) and usually say that x² + 9 "does not factor." Confirm that this is true over the real numbers. Then use (3i)² = -9 to rewrite x² + 9 as x² - (3i)², a difference of squares, so x² + 9 = (x + 3i)(x - 3i). Tell students that today they will see which familiar identities keep working when the numbers inside them are complex.
Direct Instruction20 minutes
A polynomial identity such as A² - B² = (A + B)(A - B) is true for every value of A and B, because it only uses the commutative, associative and distributive properties. Complex numbers follow the same properties, so the identity stays true when B is a complex number. The only new fact is i² = -1. Use Diagram 1 to show three identities and their complex versions:
Difference of squares: with B = bi, B² = -b², so A² + b² = (A + bi)(A - bi). A sum of two squares now factors.
Square of a binomial: (A + bi)² = A² + 2Abi + (bi)² = A² + 2Abi - b². The last term changes sign.
Product of conjugate factors: (x - (p + qi))(x - (p - qi)) = (x - p)² + q² = x² - 2px + p² + q². A quadratic with real coefficients that is a shifted sum of squares factors into conjugate linear factors.
Check every factorization by multiplying it back out, using i² = -1.
Official example: sum of squares
Rewrite x² + 4 as a product of linear factors. Since (2i)² = -4, x² + 4 = x² - (2i)².
Factor x² + 2x + 10. Completing the square gives (x + 1)² + 9, a sum of squares with A = x + 1.
Equation: x² + 2x + 10 = (x + 1 + 3i)(x + 1 - 3i)
Use Diagram 2 to multiply the official example back out in a grid: the two ix terms cancel and -4i² becomes +4. Point out the contrast with the fourth example: (x + 3i)² is not x² + 9, because the cross terms do not cancel when both factors are the same.
Guided Practice15 minutes
Pairs work through five items. For each one, they first name the identity they will use and write what plays the role of A and B.
Guided practice items with the identity and the result
Expression
Identity
Result
x² + 81
Difference of squares, B = 9i
(x + 9i)(x - 9i)
4x² + 1
Difference of squares, A = 2x, B = i
(2x + i)(2x - i)
x² + 3
Difference of squares, B = i√3
(x + i√3)(x - i√3)
(x - 2i)²
Square of a binomial
x² - 4ix - 4
x² - 4x + 13
Complete the square: (x - 2)² + 9
(x - 2 + 3i)(x - 2 - 3i)
Listen for students who write (x + 9)(x - 9)i or (x + 9i)². Ask them to multiply their answer back out and compare it with the original expression.
Independent Practice10-15 minutes
Students work alone on six items and check each by multiplying: x² + 225 = (x + 15i)(x - 15i), 25x² + 16 = (5x + 4i)(5x - 4i), x² + 11 = (x + i√11)(x - i√11), (2x + i)² = 4x² + 4ix - 1, x² + 6x + 10 = (x + 3 + i)(x + 3 - i), and x⁴ - 16 = (x - 2)(x + 2)(x - 2i)(x + 2i). For the last one, students first use the difference of squares on x⁴ - 16 = (x² - 4)(x² + 4) and then use it again on each factor.
Closure5 minutes
Exit ticket: (1) Factor x² + 100 into linear factors. (Answer: (x + 10i)(x - 10i).) (2) A student writes x² + 49 = (x + 7i)². Explain the error in one sentence. (Answer: (x + 7i)² = x² + 14ix - 49; the factors must be conjugates.) (3) Name the real identity that you extended in question 1.
Differentiation Strategies
For Struggling Students
Give a template A² - B² = (A + B)(A - B) with blank boxes for A and B, and have students fill them in before writing the factors
Use the multiplication grid from Diagram 2 for every check, so the cancellation of the middle terms is visible
Start with sums of squares of perfect squares, such as x² + 1 and x² + 25, before irrational cases like x² + 7
For Advanced Students
Ask students to factor x⁴ + 4 into two real quadratics and then into four complex linear factors
Ask whether x² + 2ix - 1 is a perfect square, and which identity shows it
Ask students to show that (a² + b²)(c² + d²) is a sum of two squares by factoring each part with complex numbers and regrouping the factors
Assessment Guidance
What to Look For
Look for students who can name the real identity behind each complex factorization, not only produce the factors. A correct answer to a sum of squares has two conjugate factors, such as (5x + 4i)(5x - 4i), never a single squared factor. When students expand squares like (2x + i)², check that the last term is negative. Ask every student to multiply at least one factorization back out: this check is the fastest way to catch sign errors with i² = -1.
02
Classroom Activities
3 Activities
1
Derive the Conjugate Product Identity
15 minGroups of 3
Each group receives 3 identity cards and derives a general identity for the product of two conjugate linear factors, then tests it on three cases. This shows students how a new complex identity comes from an old real one.
Procedure
Card 1: expand (x - (p + qi))(x - (p - qi)) by regrouping it as ((x - p) - qi)((x - p) + qi)
Card 2: use the difference of squares to reach (x - p)² - (qi)² = (x - p)² + q² = x² - 2px + p² + q²
Card 3: test the identity with (p, q) = (1, 2), (-2, 1) and (0, 3); the group should get x² - 2x + 5, x² + 4x + 5 and x² + 9
Discussion Questions
Why are all the coefficients of the result real, even though the factors contain i?
What does the identity say about a quadratic x² + bx + c whose factors are conjugates?
Where in your work did you use i² = -1?
Modification for Distance Learning
Share the three cards as slides. Each student in a breakout group completes one card, and the group checks that the three results fit together.
2
Factor Dominoes
20 minPairs
Pairs cut out 8 dominoes. Each domino has an expanded expression on one half and a factored or simplified form on the other half, and the halves of different dominoes match. Pairs arrange the dominoes into a closed loop.
The 8 Matching Pairs
x² + 1 and (x + i)(x - i)
16x² + 9 and (4x + 3i)(4x - 3i)
x² + 7 and (x + i√7)(x - i√7)
x² - 8x + 17 and (x - 4 + i)(x - 4 - i)
x² + 10x + 29 and (x + 5 + 2i)(x + 5 - 2i)
(x + 4i)² and x² + 8ix - 16
(1 + 5i)(1 - 5i) and 26
x⁴ - 1 and (x - 1)(x + 1)(x - i)(x + i)
Procedure
The teacher prints each domino with the left half of one pair and the right half of the next pair, so that the 8 dominoes form one loop
Pairs connect dominoes only after writing the multiplication that proves the match
Pairs label each match with the identity it uses
Challenge Variation
Pairs add two dominoes of their own, one using a sum of squares with an irrational b and one using the square of a binomial, and insert them into the loop.
3
Sums of Two Squares as Products
15 minPairs
Students apply the identity p² + q² = (p + qi)(p - qi) to whole numbers. The same identity that factors x² + 4 factors numbers that are sums of two squares into complex conjugates.
Procedure
Write each number as a sum of two squares, then as a product of conjugates: 10 = 3² + 1² = (3 + i)(3 - i), 13 = (3 + 2i)(3 - 2i), 29 = (5 + 2i)(5 - 2i), 41 = (5 + 4i)(5 - 4i)
Find two different ways to write 50: 50 = (7 + i)(7 - i) = (5 + 5i)(5 - 5i)
Check every product by multiplying
Discussion Questions
Why can 3 not be written as (p + qi)(p - qi) with whole numbers p and q?
How is 10 = (3 + i)(3 - i) the same identity as x² + 1 = (x + i)(x - i) with x = 3?
Challenge Variation
Challenge (beyond the standard): multiply (3 + i)(2 + i) and (3 - i)(2 - i), and use the results to write 10 · 5 = 50 as a sum of two squares. Ask students to explain why this always works.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Three Real Identities and Their Complex Versions
Each row starts with an identity that is true for all real A and B. Replacing B with bi, where b is real, uses (bi)² = -b² and gives the complex version in the middle column. The right column shows an example from this lesson for each row, including the official example x² + 4 = (x + 2i)(x - 2i).
Diagram 2: A Product Grid for the Official Example
The grid multiplies each term of x + 2i by each term of x - 2i. The two middle products 2ix and -2ix cancel, and -4i² = +4 because i² = -1, so (x + 2i)(x - 2i) = x² + 4. The grid organizes the products; its cells are not drawn to scale.
04
Homework Assignment
~30 min
HSN.CN.C.8 Homework: Polynomial Identities with Complex Numbers
Directions: Show all work. For each factorization, name the identity you used and check your answer by multiplying the factors back out. Use i² = -1.
Part 1: Sums of Squares (Problems 1-2)
Factor each expression into linear factors over the complex numbers: (a) x² + 144 (b) 49x² + 4 (c) x² + 13
Write 36x² + 25 as a product of two linear factors. Then multiply your factors to show that the ix terms cancel.
Part 2: Other Identities (Problems 3-4)
Use (A + B)² = A² + 2AB + B² or (A - B)² = A² - 2AB + B² to expand (a) (x + 9i)² and (b) (2x - 3i)².
Complete the square and then factor each quadratic into linear factors: (a) x² - 10x + 26 (b) x² + 12x + 40
Part 3: Degree 4 and Numbers (Problems 5-6)
Factor completely over the complex numbers: (a) x⁴ - 625 (b) x⁴ + 5x² + 4. For (b), first factor as a product of two quadratics in x².
Use the identity p² + q² = (p + qi)(p - qi) to write 53 as a product of two conjugate complex numbers with whole-number parts. Then find two different such products that equal 65.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Identity
Correct identity named for every problem
Identity named for some problems
No identity named
Factoring
All factors correct and linear, in conjugate pairs where needed
One factor wrong or not fully factored
Most factorizations wrong
Use of i² = -1
Signs correct in every expansion
One sign error
i² treated as 1 or ignored
Checking
Every factorization multiplied back out
Some checks shown
No checks
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which is a factorization of x² + 64 over the complex numbers?
Answer: C
x² + 64 = x² - (8i)² = (x + 8i)(x - 8i), the difference of squares with B = 8i. Choice A multiplies to x² - 64. Choice B expands to x² + 16ix - 64 because its middle terms do not cancel.
Question 2 of 20 · Multiple Choice
Which is a factorization of 4x² + 9?
Answer: A
Here A = 2x and B = 3i, so 4x² + 9 = (2x)² - (3i)² = (2x + 3i)(2x - 3i). Choice B squares to 16x² + 81, not 4x² + 9. Choice C gives 4x² - 9.
Question 3 of 20 · Multiple Choice
Which is a factorization of x² + 6?
Answer: B
Since 6 = (√6)², x² + 6 = x² - (i√6)² = (x + i√6)(x - i√6). Choice A multiplies to x² + 36. Choice D multiplies to x² - 6.
Question 4 of 20 · Multiple Choice
Expand (x + 6i)(x - 6i).
Answer: D
(x + 6i)(x - 6i) = x² - (6i)² = x² - 36i² = x² + 36. Choice A forgets that i² = -1. Choice B is the expansion of (x + 6i)², a different product.
Question 5 of 20 · Multiple Choice
Which is the expansion of (x - 5i)²?
Answer: C
(A - B)² = A² - 2AB + B² with B = 5i: x² - 10ix + (5i)² = x² - 10ix - 25. Choice D uses 25 instead of (5i)² = -25. Choices A and B drop the middle term 2AB.
Question 6 of 20 · Multiple Choice
Which is a factorization of x² - 2x + 2?
Answer: A
Complete the square: x² - 2x + 2 = (x - 1)² + 1 = (x - 1 + i)(x - 1 - i). Choice B has the wrong sign for the real part and gives x² + 2x + 2. Choice C gives (x - 1)² + 4 = x² - 2x + 5.
Question 7 of 20 · Multiple Choice
Which product equals x² + 8x + 20?
Answer: D
x² + 8x + 20 = (x + 4)² + 4 = (x + 4)² - (2i)², so the factors are (x + 4 + 2i)(x + 4 - 2i). Choice A gives (x + 4)² + 16. Choice B has the wrong sign for the real part.
Question 8 of 20 · Multiple Choice
Simplify (6 + i)(6 - i).
Answer: B
By the identity (p + qi)(p - qi) = p² + q², the product is 36 + 1 = 37. Choice A treats i² as +1. Choice D has the wrong sign on the last term: the product is 36 - 6i + 6i - i² = 36 - i² = 37, not 36 + i², which equals 35.
Question 9 of 20 · Multiple Choice
A student writes x² + 25 = (x + 5i)². What is correct?
Answer: C
A sum of squares factors into conjugates, not a square. Squaring gives (x + 5i)² = x² + 10ix - 25, which is not x² + 25. Choice D is true only over the real numbers.
Question 10 of 20 · Multiple Choice
Which is the complete factorization of x⁴ - 81 over the complex numbers?
Answer: A
x⁴ - 81 = (x² - 9)(x² + 9), then x² - 9 = (x - 3)(x + 3) and x² + 9 = (x - 3i)(x + 3i). Choice B uses 9i instead of 3i, so its last two factors multiply to x² + 81. Choice D multiplies to (x² - 9)², a different polynomial.
Question 11 of 20 · Multiple Choice
Which is a factorization of x⁴ + 10x² + 9 into linear factors?
Answer: D
x⁴ + 10x² + 9 = (x² + 1)(x² + 9), and each sum of squares factors: (x + i)(x - i)(x + 3i)(x - 3i). Choice A gives (x² - 1)(x² - 9). Choice C uses 9i instead of 3i, so its second pair multiplies to x² + 81.
Question 12 of 20 · Multiple Choice
Which statement about x² + 10 is true?
Answer: B
No real number squares to -10, so there are no real linear factors. Over the complex numbers, 10 = (√10)² gives (x + i√10)(x - i√10). Choice A multiplies to x² + 100, and choice C to x² - 10.
Question 13 of 20 · Multiple Choice
Which expression equals (x + 1 + 4i)(x + 1 - 4i)?
Answer: A
Group as ((x + 1) + 4i)((x + 1) - 4i) = (x + 1)² - (4i)² = x² + 2x + 1 + 16 = x² + 2x + 17. Choice B forgets that (4i)² = -16. Choice D drops the 2x from (x + 1)².
Question 14 of 20 · Multiple Choice
For which value of k is x² + k = (x + 3i√2)(x - 3i√2)?
Answer: C
The product is x² - (3i√2)² = x² - 9 · 2 · i² = x² + 18, so k = 18. Choice A treats i² as +1. Choice B forgets to square the 3.
Show that x² + 12 = (x + 2i√3)(x - 2i√3) and name the identity you extended.
Multiply: x² - (2i√3)² = x² - 4 · 3 · i² = x² + 12. The identity is the difference of squares A² - B² = (A + B)(A - B) with A = x and B = 2i√3, a complex number.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSN.CN.C.8 mean?
HSN.CN.C.8 means students use familiar polynomial identities with complex numbers. The official example rewrites x² + 4 as (x + 2i)(x - 2i): the difference of squares, with B = 2i, turns a sum of squares into a product of linear factors.
Is HSN.CN.C.8 Algebra 2 or Precalculus?
It can appear in either. HSN.CN.C.8 is a (+) standard, so it belongs to the additional mathematics for students who take advanced courses. Many schools meet it in Algebra II right after complex numbers are introduced, and again in Precalculus when polynomials are factored completely.
What does the (+) in front of HSN.CN.C.8 mean?
The (+) marks standards that Common Core describes as additional mathematics that students should learn in order to take advanced courses such as calculus. They are not required of every student, but they are part of the high school standards.
I was taught that a sum of squares cannot be factored. Is that wrong?
It is true over the real numbers and false over the complex numbers. x² + 1 has no real linear factors because no real number squares to -1. With i, x² + 1 = (x + i)(x - i). Teachers should say "does not factor over the reals" rather than "does not factor."
Why do polynomial identities still work with complex numbers?
An identity like A² - B² = (A + B)(A - B) is proved using only the commutative, associative and distributive properties. Complex numbers obey the same properties, so the proof works word for word when A or B is complex. The only new fact students need is i² = -1.
What are common mistakes with complex identities?
Three errors come up often: writing a sum of squares as a single square, such as x² + 16 = (x + 4i)²; forgetting that (bi)² = -b², which flips the sign of the last term; and dropping the middle term 2AB when squaring a binomial. Multiplying the answer back out catches all three.
How does HSN.CN.C.8 connect to solving quadratic equations?
Each linear factor gives a solution. Because x² + 4 = (x + 2i)(x - 2i), the equation x² + 4 = 0 has the solutions ±2i. This links the identity work here to HSN.CN.C.7, which asks students to solve quadratic equations with complex solutions.
How does HSN.CN.C.8 connect to the Fundamental Theorem of Algebra?
The Fundamental Theorem of Algebra (HSN.CN.C.9) says that a polynomial of degree n has n complex zeros, counted with multiplicity, so it splits into n linear factors over the complex numbers. The identities in this standard are how students actually find those factors for quadratics and for polynomials like x⁴ - 16.
Can whole numbers be factored with complex numbers too?
Yes, when they are sums of two squares. The identity p² + q² = (p + qi)(p - qi) gives 5 = (2 + i)(2 - i) and 17 = (4 + i)(4 - i). This is a first look at factoring in the Gaussian integers, a topic from number theory.
How can students check a complex factorization?
Multiply the factors and use i² = -1. For x² + 20 = (x + 2i√5)(x - 2i√5), the product is x² - (2i√5)² = x² - 20i² = x² + 20. A grid like Diagram 2 helps students see that the two middle terms cancel.
07
Related Standards
5 standards
These standards connect to HSN.CN.C.8: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSN.CN.A.2Prerequisite
Add, subtract and multiply complex numbers using i² = -1 and number properties
