HSN.CN.C.9Common CoreMathNumber and QuantityGrades 9-12
HSN.CN.C.9: The Fundamental Theorem of Algebra
In plain English: HSN.CN.C.9 is an advanced (+) Common Core number and quantity standard, usually taught in Algebra II or Precalculus. Students learn the Fundamental Theorem of Algebra: every polynomial of degree n ≥ 1 has a complex zero, so it has exactly n zeros counted with multiplicity. They show it is true for quadratics by using the quadratic formula in every discriminant case.
(+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.
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.9 or N-CN.9 · Official standard
Students have solved quadratic equations whose solutions are real, repeated or complex. This lesson names the pattern behind that work. The Fundamental Theorem of Algebra says that every polynomial of degree n ≥ 1 has at least one complex zero. Factoring out one zero at a time then shows that the polynomial has exactly n zeros, counted with multiplicity, all of them complex numbers (real numbers included).
Students then prove the theorem for quadratics. Completing the square on ax² + bx + c gives the quadratic formula, and the formula produces two complex zeros whether the discriminant is positive, zero or negative. Graphs show why real zeros can go missing on the x-axis while the complex zeros are still there.
Learning Objectives
By the end of this lesson, students will be able to:
State the Fundamental Theorem of Algebra and explain what "counted with multiplicity" means
Use the theorem to predict how many complex zeros a polynomial has from its degree
Show that every quadratic polynomial has exactly two complex zeros, counted with multiplicity, by completing the square
Classify the zeros of a quadratic with real coefficients using the discriminant and connect each case to its graph
Prior Knowledge Required
Students should already be comfortable with:
Solving quadratic equations with the quadratic formula and completing the square HSA.REI.B.4
Writing complex solutions in the form a + bi HSN.CN.C.7
Factoring sums of squares with complex numbers HSN.CN.C.8
Connecting zeros and factors of a polynomial HSA.APR.B.2
Start with three equations that differ only in their sign or constant.
Warm-Up Prompt
"How many solutions does each equation have: x² - 9 = 0, x² - 6x + 9 = 0, x² + 9 = 0? Does your answer change if you allow complex numbers?"
Students should find 3 and -3; only 3, because x² - 6x + 9 = (x - 3)²; and no real solutions but 3i and -3i. Ask how to make the count come out the same every time. Counting 3 twice for (x - 3)² and allowing complex numbers gives two solutions in all three cases. That is the idea of the lesson.
Direct Instruction15 minutes
Part 1: The theorem. The Fundamental Theorem of Algebra: every polynomial of degree n ≥ 1 with complex coefficients has at least one complex zero. If r is a zero, then x - r is a factor, so p(x) = (x - r)q(x) with q of degree n - 1. Repeating the argument on q splits p into n linear factors, so p has exactly n zeros counted with multiplicity. The zero r has multiplicity k when (x - r)k is a factor but (x - r)k+1 is not.
Part 2: A proof for quadratics. For ax² + bx + c with real coefficients and a ≠ 0, complete the square:
Divide by a: ax² + bx + c = 0 has the same solutions as x² + (b/a)x + c/a = 0.
Complete the square: (x + b/(2a))² = (b² - 4ac)/(4a²). Call D = b² - 4ac the discriminant.
Take square roots in the complex numbers: if D ≥ 0, √D is real; if D < 0, √D = i√(-D). Either way a square root exists, so x = (-b ± √D)/(2a).
Count: the formula gives two zeros r₁ and r₂. They are equal exactly when D = 0, which is a zero of multiplicity 2.
Factor: ax² + bx + c = a(x - r₁)(x - r₂), a product of exactly two linear factors. So every quadratic has exactly 2 complex zeros, counted with multiplicity.
Two real zeros (positive discriminant)
Find the zeros of x² - 2x - 15 and check the count.
Equation: D = 4 + 60 = 64, x = (2 ± 8)/2, so x = 5 or x = -3; x² - 2x - 15 = (x - 5)(x + 3)
How many zeros does p(x) = x³ - 2x² + 3x - 6 have? Factor by grouping and find them.
Equation: p(x) = (x - 2)(x² + 3), zeros 2, i√3 and -i√3: three zeros for degree 3
Multiplicity
List the zeros of p(x) = (x - 1)²(x² + 25) with their multiplicities.
Equation: 1 (multiplicity 2), 5i and -5i: 2 + 1 + 1 = 4, the degree
Guided Practice15 minutes
Pairs fill in the table. For each quadratic they compute the discriminant first, predict the kind of zeros, and then solve to confirm that the count is 2.
Guided practice
Quadratic
Discriminant
Zeros
x² + 6x + 5
36 - 20 = 16
-1 and -5
9x² - 30x + 25
900 - 900 = 0
5/3, multiplicity 2
x² + 2x + 5
4 - 20 = -16
-1 + 2i and -1 - 2i
2x² - 3x + 4
9 - 32 = -23
(3 + i√23)/4 and (3 - i√23)/4
Ask pairs to sketch each graph on a calculator. The first crosses the x-axis twice, the second touches it once, and the last two never meet it. The theorem still counts two zeros for each: the graph shows only the real ones.
Independent Practice15 minutes
Students work alone. For each quadratic, find the discriminant, then solve: (1) x² - 7x + 12 (3 and 4); (2) x² + 10x + 25 (-5, multiplicity 2); (3) x² - 6x + 10 (3 + i and 3 - i); (4) 3x² + 2x + 1 ((-1 + i√2)/3 and (-1 - i√2)/3). Then (5) state how many zeros x⁵ - x has and find them (5 zeros: 0, 1, -1, i and -i). As a stretch item, write a quadratic with real coefficients whose zeros are 1 + 4i and 1 - 4i (x² - 2x + 17).
Closure5 minutes
Exit ticket: (1) Complete the sentence: a polynomial of degree n ≥ 1 has exactly ___ complex zeros, counted with multiplicity. (Answer: n.) (2) Solve x² + 4x + 8 = 0 and explain how the answer agrees with the theorem. (Answer: -2 + 2i and -2 - 2i, two zeros for degree 2.) (3) Explain why a quadratic cannot have three different zeros.
Differentiation Strategies
For Struggling Students
Give a three-column organizer for D > 0, D = 0 and D < 0 with a sample graph and a sample pair of zeros in each column
Write each polynomial in factored form first, so students can count the factors before they count the zeros
Use the phrase "count with multiplicity" with a concrete picture: (x - 2)² is the factor x - 2 used twice
For Advanced Students
Ask students to prove that if r₁ and r₂ are the zeros of ax² + bx + c, then r₁ + r₂ = -b/a and r₁r₂ = c/a
Ask students to explain why a polynomial with real coefficients and odd degree must have at least one real zero
Ask students to find the two square roots of i and then solve x² = i, a quadratic with a complex coefficient (going further than the standard)
Assessment Guidance
What to Look For
Students should connect degree and zero count every time and should count repeated zeros by multiplicity. In the quadratic proof, look for the step where a square root of a negative discriminant is written as i√(-D), and for the factored form a(x - r₁)(x - r₂) at the end. Watch for students who say a quadratic with a negative discriminant has no solutions without saying "no real solutions".
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Classroom Activities
3 Activities
1
Discriminant Sort
15 minGroups of 3-4
Groups sort 9 quadratic cards into three columns on a mat: two different real zeros, one real zero of multiplicity 2, and two nonreal zeros. Then they solve every card to confirm that each quadratic has exactly two zeros, counted with multiplicity.
The 9 Cards
x² - x - 6 (3 and -2)
x² - 5 (√5 and -√5)
2x² + 7x + 3 (-1/2 and -3)
x² - 8x + 16 (4, twice)
25x² + 10x + 1 (-1/5, twice)
x² + 14x + 49 (-7, twice)
x² + 16 (4i and -4i)
x² - 2x + 26 (1 + 5i and 1 - 5i)
x² + x + 1 (-1/2 + (√3/2)i and -1/2 - (√3/2)i)
Procedure
Compute each discriminant and place the card in a column before solving
Split the solving among group members, then check each other's zeros
Graph one card from each column and compare the x-intercepts with the zeros
Discussion Questions
Every card has 2 zeros counted with multiplicity. Which column needed the complex numbers to make that true?
Why do the nonreal zeros always come as a + bi and a - bi when the coefficients are real?
2
Build the Proof
15 minPairs
Pairs get 6 strips, each with one step of the proof that every quadratic ax² + bx + c with real coefficients and a ≠ 0 has two complex zeros counted with multiplicity. They put the strips in order and write the reason for each step.
The 6 Strips
Divide by a: x² + (b/a)x + c/a = 0
Move the constant: x² + (b/a)x = -c/a
Add b²/(4a²) to both sides: (x + b/(2a))² = (b² - 4ac)/(4a²)
Every real number D = b² - 4ac has a square root in the complex numbers: √D if D ≥ 0 and i√(-D) if D < 0
So x + b/(2a) = ±√D/(2a), and x = (-b + √D)/(2a) or x = (-b - √D)/(2a)
Therefore ax² + bx + c = a(x - r₁)(x - r₂) with two zeros, equal only when D = 0
Procedure
Order the strips and justify each step in the margin
Test the finished proof on 2x² + 2x + 5: D = 4 - 40 = -36, so the zeros are -1/2 + (3/2)i and -1/2 - (3/2)i
Check the last strip by expanding a(x - r₁)(x - r₂) for this example
Modification for Distance Learning
Put the strips on a shared slide in random order. Pairs drag them into place and add a comment with the reason for each step.
3
Degree Detective
15 minSmall groups
Groups get 4 polynomials in factored form. They predict the number of zeros from the degree, list every zero with its multiplicity, and check that the multiplicities add up to the degree.
Record each zero and its multiplicity in a table and add the multiplicities
Graph each polynomial and mark which zeros appear as x-intercepts
Challenge Variation
Make a polynomial of degree 5 with real coefficients that has exactly one real zero. Explain why no polynomial with real coefficients and degree 5 can have zero real zeros.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Three Quadratics, Three Discriminant Cases
The graphs of y = x² - 4x + 3, y = x² - 4x + 4 and y = x² - 4x + 5 are the same parabola shifted up 1 unit at a time. The first crosses the x-axis at 1 and 3, the second touches it at 2, and the third stays above it, with lowest point (2, 1), yet its zeros 2 + i and 2 - i still exist. Drawn to scale, with grid lines 1 unit apart on both axes (the horizontal unit is drawn wider than the vertical unit).
Diagram 2: The Zeros on the Complex Plane
The zeros of the three quadratics from Diagram 1 plotted as complex numbers. Two real zeros, one real zero of multiplicity 2, and the conjugate pair 2 ± i off the real axis: each quadratic has exactly 2 zeros counted with multiplicity. Drawn to scale, one grid square is one unit.
04
Homework Assignment
~30 min
HSN.CN.C.9 Homework: The Fundamental Theorem of Algebra
Directions: Show your work. When you state how many zeros a polynomial has, name the Fundamental Theorem of Algebra and count with multiplicity. Write complex zeros in the form a + bi.
Part 1: Knowing the Theorem (Problems 1-3)
Use the Fundamental Theorem of Algebra to state how many complex zeros, counted with multiplicity, each polynomial has: (a) 4x⁵ - x² + 7 (b) (x + 2)³(x - 6) (c) 3x - 12. For (b) and (c), also list the zeros.
Let p(x) = x³ + 36x. Factor p, find all of its zeros, and show that the number of zeros matches the degree.
A classmate says: "x² + 49 = 0 has no solutions, so the Fundamental Theorem of Algebra does not work for x² + 49." Explain the mistake and find the solutions.
Part 2: Proving It for Quadratics (Problems 4-6)
For each quadratic, find the discriminant, find the zeros and write the quadratic as a product of linear factors: (a) x² + 3x - 10 (b) 16x² - 8x + 1 (c) x² - 10x + 29.
Complete the square to show that x² + bx + c = (x + b/2)² - (b² - 4c)/4. Then explain why x² + bx + c always has two complex zeros counted with multiplicity, in each of the cases b² - 4c > 0, b² - 4c = 0 and b² - 4c < 0.
(a) Write a quadratic with real coefficients and leading coefficient 1 whose zeros are 3 + 2i and 3 - 2i. (b) Use the discriminant to find every real number k for which x² + 2x + k has two nonreal zeros.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Stating the theorem
Zero counts match the degree, with multiplicity counted correctly
Count correct but multiplicity ignored once
Counts only real zeros
Solving quadratics
Discriminant, zeros and factored form all correct
One arithmetic error
Method missing or incorrect
Proof for quadratics
Completing the square correct and all three cases explained
Algebra correct, one case missing
No proof
Reasoning
Clear correction of the classmate and a correct range of k
Partial reasoning
No explanation
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 statement is the Fundamental Theorem of Algebra?
Answer: B
The theorem guarantees a complex zero for every nonconstant polynomial, and factoring then gives exactly n zeros counted with multiplicity. Choice A says real, which fails for x² + 1. Choice D fails for the same polynomial. Choice C fails for x² - 2.
Question 2 of 20 · Multiple Choice
How many complex zeros, counted with multiplicity, does p(x) = 2x⁶ - 3x⁴ + x - 9 have?
Answer: A
The degree is 6, so the theorem gives exactly 6 complex zeros counted with multiplicity. Choice B uses the second exponent. Choice C uses the leading coefficient, and choice D uses the constant term.
Question 3 of 20 · Multiple Choice
What are the zeros of x² + 6x + 13?
Answer: C
D = 36 - 52 = -16, so x = (-6 ± 4i)/2 = -3 ± 2i. Choice A drops the sign of -b. Choice B forgets to divide √(-16) = 4i by 2. Choice D confuses no real zeros with no zeros: the theorem guarantees two complex zeros.
Question 4 of 20 · Multiple Choice
Which describes the zeros of 9x² - 12x + 4?
Answer: D
D = 144 - 144 = 0, and 9x² - 12x + 4 = (3x - 2)², so 2/3 is a zero of multiplicity 2. Counted with multiplicity, that is 2 zeros. Choice A would need a negative discriminant, and choice B a positive one.
Question 5 of 20 · Multiple Choice
For p(x) = (x - 4)²(x + 1)(x² + 1), how many distinct zeros does p have, and how many zeros counted with multiplicity?
Answer: B
The zeros are 4 (multiplicity 2), -1, i and -i: 4 distinct zeros, and 2 + 1 + 1 + 1 = 5 with multiplicity, which is the degree. Choice D counts only the real zeros. Choice C counts 4 twice as two different zeros.
Question 6 of 20 · Multiple Choice
What are the zeros of x² - 2x + 5?
Answer: A
D = 4 - 20 = -16, so x = (2 ± 4i)/2 = 1 ± 2i. Choice B takes -b as -2. Choice C does not divide 4i by 2. Choice D adds 4 + 20 in the discriminant, a sign error with c.
Question 7 of 20 · Multiple Choice
A polynomial of degree 7 has the zeros 0, 2 and -2, each of multiplicity 1. How many more zeros, counted with multiplicity, does it have?
Answer: C
By the theorem there are 7 zeros counted with multiplicity. Three are known, so 7 - 3 = 4 remain; they may be nonreal or repeated. Choice A counts the known zeros instead. Choice D assumes the listed zeros are all of them.
Question 8 of 20 · Multiple Choice
Which quadratic has the zeros 4 + i and 4 - i?
Answer: D
(x - (4 + i))(x - (4 - i)) = (x - 4)² - i² = x² - 8x + 16 + 1 = x² - 8x + 17. Choice A has zeros -4 ± i. Choice B treats i² as 1, giving 16 - 1. Choice C has the wrong sign on the constant, and its zeros are real.
Question 9 of 20 · Multiple Choice
Why does the equation x² + 5 = 0 not contradict the Fundamental Theorem of Algebra?
Answer: A
x² = -5 gives x = ±i√5, two complex zeros for a degree 2 polynomial, as the theorem says. Choice B is false: the theorem covers every degree n ≥ 1. Choice C is false: the theorem is about complex zeros. Choice D solves x² - 5 = 0 instead.
Question 10 of 20 · Multiple Choice
Completing the square gives x² - 8x + 25 = (x - 4)² + 9. What are the zeros?
Answer: B
(x - 4)² = -9, so x - 4 = ±3i and x = 4 ± 3i. Choice A forgets the square root of 9. Choice C gets the sign of the shift wrong. Choice D takes √(-9) as 3, which gives 1 and 7, and neither is a zero.
Question 11 of 20 · Multiple Choice
A quadratic with real coefficients has a negative discriminant. What do you know about its zeros?
Answer: C
With D < 0, the formula gives x = -b/(2a) ± (√(-D)/(2a))i, a conjugate pair. Choice A describes D = 0. Choice B ignores the complex numbers. Choice D describes some cases with D > 0.
Question 12 of 20 · Multiple Choice
Which polynomial has exactly 3 complex zeros, counted with multiplicity?
Answer: D
The degree of 5 + 3x² - x³ is 3, the highest exponent, even though that term is written last. So it has 3 zeros counted with multiplicity. Choice A has degree 4 and choice C has degree 6. Choice B has degree 1, and its coefficient 3 is not the degree.
Question 13 of 20 · Multiple Choice
A student says 4x² + 4x + 1 = 0 has only one solution, so the Fundamental Theorem of Algebra fails for it. What is correct?
Answer: A
D = 16 - 16 = 0 and 4x² + 4x + 1 = (2x + 1)². The factor 2x + 1 appears twice, so -1/2 has multiplicity 2 and the count is 2. Choice B is wrong because x = 1/2 gives 1 + 2 + 1 = 4, not 0. Choice D is false: the theorem covers every nonconstant polynomial.
Question 14 of 20 · Multiple Choice
A cubic polynomial with real coefficients has exactly one real zero, and that zero has multiplicity 1. How many nonreal zeros does it have?
Answer: C
A cubic has 3 zeros counted with multiplicity. One is real with multiplicity 1, so the other 2 are nonreal (they form a conjugate pair). Choice A would leave only one zero in all. Choice B is impossible because nonreal zeros of a polynomial with real coefficients come in pairs.
Question 15 of 20 · Short Answer
Find the zeros of 2x² + 4x + 5 and explain how they agree with the Fundamental Theorem of Algebra.
D = 16 - 40 = -24, so x = (-4 ± √(-24))/4 = (-4 ± 2i√6)/4 = -1 ± (√6/2)i. The polynomial has degree 2 and exactly 2 complex zeros, as the theorem says.
Question 16 of 20 · Short Answer
Find all zeros of p(x) = x³ + x² + 9x + 9 and check the count against the degree.
Group: x²(x + 1) + 9(x + 1) = (x + 1)(x² + 9). The zeros are -1, 3i and -3i: three zeros for degree 3.
Question 17 of 20 · Short Answer
Write a quadratic with real coefficients and leading coefficient 1 whose zeros are -2 + 5i and -2 - 5i.
For which values of k does x² + kx + 36 have a zero of multiplicity 2? Give the zero for each value.
A double zero needs D = k² - 144 = 0, so k = 12 or k = -12. For k = 12, x² + 12x + 36 = (x + 6)² and the zero is -6. For k = -12, x² - 12x + 36 = (x - 6)² and the zero is 6.
Question 19 of 20 · Short Answer
The graph of y = x⁴ + 1 never touches the x-axis. Explain why x⁴ + 1 still has 4 complex zeros, and say whether any of them are real.
x⁴ + 1 has degree 4, so the theorem gives exactly 4 complex zeros counted with multiplicity. For every real x, x⁴ + 1 ≥ 1 > 0, so none of them are real. (They are √2/2 + (√2/2)i, √2/2 - (√2/2)i, -√2/2 + (√2/2)i and -√2/2 - (√2/2)i.) The graph shows only real zeros, so it cannot show these.
Question 20 of 20 · Short Answer
Write 2x² + 8x + 10 as a product of a constant and two linear factors over the complex numbers.
2x² + 8x + 10 = 2(x² + 4x + 5). For x² + 4x + 5, D = 16 - 20 = -4 and x = (-4 ± 2i)/2 = -2 ± i. So 2x² + 8x + 10 = 2(x + 2 - i)(x + 2 + i), two linear factors, as the theorem predicts.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSN.CN.C.9 mean?
It means students know the Fundamental Theorem of Algebra and can show it is true for quadratics. The theorem says every polynomial of degree n ≥ 1 has a complex zero, and therefore exactly n zeros counted with multiplicity. For quadratics, students use completing the square or the quadratic formula to show there are always two.
What is the Fundamental Theorem of Algebra in simple terms?
Every nonconstant polynomial equation has a solution if you allow complex numbers. A polynomial of degree n has exactly n solutions when repeated ones are counted as many times as their factor appears. For example, x³ - x has degree 3 and the three zeros 0, 1 and -1.
Is HSN.CN.C.9 Algebra 2 or Precalculus?
It is usually taught in Algebra II, with complex solutions of quadratics and polynomial functions, and revisited in Precalculus. The (+) marks it as additional mathematics that Common Core describes for students taking advanced courses.
Do students have to prove the Fundamental Theorem of Algebra?
No, only for quadratic polynomials. The standard asks students to know the theorem and to show it is true for quadratics. General proofs use ideas from college mathematics, such as complex analysis, so high school courses state the theorem and prove only the degree 2 case.
What does "counted with multiplicity" mean?
A zero is counted once for each time its factor appears. In (x - 2)³(x + 1), the zero 2 has multiplicity 3 and -1 has multiplicity 1, so the polynomial has 4 zeros counted with multiplicity, matching its degree, but only 2 distinct zeros.
How does the quadratic formula show the theorem is true for quadratics?
The formula x = (-b ± √(b² - 4ac))/(2a) always gives two complex numbers. When b² - 4ac is negative, its square root is an imaginary number, so the zeros are still complex numbers. When b² - 4ac = 0, the two zeros are equal and form one zero of multiplicity 2. So every quadratic factors as a(x - r₁)(x - r₂).
Does the Fundamental Theorem of Algebra tell you how to find the zeros?
No. It guarantees that the zeros exist and tells you how many there are, but it gives no method for finding them. For quadratics the quadratic formula finds them. For polynomials of degree 5 and higher there is no general formula using radicals, so factoring, graphing and numerical methods are used.
Why does the theorem need complex numbers?
Because some polynomials have no real zeros. The graph of x² + 1 never crosses the x-axis, so x² + 1 has no real zeros, but it has the complex zeros i and -i. With only real numbers the count of zeros would change from one polynomial to the next; with complex numbers it always equals the degree.
What mistakes do students make with the Fundamental Theorem of Algebra?
A common one is saying that a quadratic with a negative discriminant has no solutions, instead of no real solutions. Others include counting a double zero only once, reading the degree from the first term written instead of the highest exponent, and thinking the theorem says all zeros are real.
How does HSN.CN.C.9 connect to other standards?
It builds on solving quadratics with complex solutions (HSN.CN.C.7) and on factoring with complex numbers (HSN.CN.C.8). It explains why the zeros and factors of a polynomial match (HSA.APR.B.2) and why graphs of polynomials show only some of their zeros (HSA.APR.B.3 and HSF.IF.C.7).
07
Related Standards
6 standards
These standards connect to HSN.CN.C.9: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSA.REI.B.4Prerequisite
Solve quadratic equations in one variable by several methods
