Algebra 2 Quiz: Extending Polynomial Identities To Complex Numbers

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Extending Polynomial Identities To Complex NumbersQuestion 1 of 20

Use the identity $a^2-b^2=(a+b)(a-b)$ , which remains valid for complex $a,b$ , to evaluate $(1+i)^2-(2-i)^2.$

3+6i

-1+2i

-3-6i

-3+6i

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Algebra 2 Quiz

Algebra 2 Quiz: Extending Polynomial Identities To Complex Numbers

Practice Extending Polynomial Identities To Complex Numbers in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Extending Polynomial Identities To Complex Numbers, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Use the identity $a^2-b^2=(a+b)(a-b)$ , which remains valid for complex $a,b$ , to evaluate $(1+i)^2-(2-i)^2.$

$3+6i$
$-1+2i$
$-3-6i$
$-3+6i$ (correct answer)

a^2 - b^2

with complex values. All polynomial identities that work for real numbers also work for complex numbers because complex numbers form an algebraically closed field—addition, subtraction, multiplication work with same properties (commutative, associative, distributive). For instance, with

a=1+i

and

b=2-i

a^2 - b^2 = (a+b)(a-b) = (3)(-1+2i) = -3+6i

, matching direct computation of

2i - (3-4i) = -3+6i

. Choice A correctly applies the polynomial identity to complex values, yielding

-3+6i

through careful arithmetic. A tempting distractor like choice B might flip the sign by miscomputing

a-b

-1-2i

instead of

-1+2i

—always double-check subtractions with imaginaries! To use difference of squares with complexes: (1) Identify

a

and

b

. (2) Compute

a+b

and

a-b

separately. (3) Multiply and simplify. This strategy works every time—keep up the great work extending identities!

Question 2

Over $\mathbb{C}$ , you can factor expressions using standard identities even when no real factorization exists. Factor $x^4+4$ completely over $\mathbb{C}$ .

$x^4+4=(x^2+2)(x^2+2)$
$x^4+4=(x^2+2x+2)(x^2-2x+2)$ (correct answer)
$x^4+4=(x^2+2ix+2)(x^2-2ix+2)$
$x^4+4=(x^2-2)(x^2+2)$

x^4 + 4

, we can use Sophie Germain's identity:

a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)

. With

a = x

and

b = 1

x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)

. Let's verify:

(x^2 + 2x + 2)(x^2 - 2x + 2) = x^4 - 2x^3 + 2x^2 + 2x^3 - 4x^2 + 4x + 2x^2 - 4x + 4 = x^4 + 4

. Perfect! Choice B correctly identifies this factorization. Choice C tries to use complex coefficients directly with

(x^2 + 2ix + 2)(x^2 - 2ix + 2)

, but this would expand to

x^4 + 4x^2 + 4

, not

x^4 + 4

—Sophie Germain's identity provides the correct real coefficient factorization that can then be further factored over complex if needed. Factoring

x^4 + 4

: (1) Recognize this fits Sophie Germain's pattern

a^4 + 4b^4

. (2) Apply the identity to get quadratic factors with real coefficients. (3) Each quadratic can be further factored over complex using the quadratic formula. The beauty: even expressions that seem "prime" often factor using clever identities!

Question 3

A student wants to verify that polynomial operations (expanding and simplifying) work the same way over complex numbers. Which option correctly computes $(2+3i)^2$ using $(a+b)^2=a^2+2ab+b^2$ with $a=2$ and $b=3i$ ?

$13-12i$
$13+12i$
$-5+12i$ (correct answer)
$-5-12i$

4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i

. Choice A correctly applies the polynomial identity to compute

-5+12i

, properly handling

(3i)^2 = -9

. Choice B forgets

i^2 = -1

, treating it as +9 for a positive real part—always simplify powers of i right away! Expanding with complex: (1) Identify a and b; (2) Compute a^2, 2ab, b^2 separately; (3) Add reals and imaginaries; (4) Verify directly. Keep up the great effort—you're mastering complex arithmetic through identities!

Question 4

Over the real numbers, $x^2+9$ does not factor. Over the complex numbers, use the difference of squares identity $u^2-v^2=(u+v)(u-v)$ by writing $x^2+9=x^2-(3i)^2$ . Which is the correct factorization over $\mathbb{C}$ ?

$(x+3)(x-3)$
$(x-3i)(x-3)$
$(x+3i)^2$
$(x+3i)(x-3i)$ (correct answer)

Explanation: This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, allowing us to factor expressions over complex that don't factor over reals. Key concept: The difference of squares identity u^2 - v^2 = (u+v)(u-v) holds for complex numbers, so x^2 +9 = x^2 - (3i)^2 since (3i)^2 = -9. Verifying: (x+3i)(x-3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 +9, perfect! Choice B correctly applies this by setting v=3i for the factorization (x+3i)(x-3i). A tempting distractor like A fails by using real numbers only, ignoring that +9 is a sum of squares factorable over complexes as (x+3i)(x-3i). Transferable strategy: For factoring sums of squares over complex: (1) Rewrite a^2 + b^2 = a^2 - (bi)^2. (2) Apply difference of squares: (a + bi)(a - bi). (3) Verify by expanding. Great work extending your factoring skills!

Question 5

Apply the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ for complex numbers with $a=1+i$ and $b=1-i$ . What is the value of $(1+i)^3-(1-i)^3$ ?

$-4i$
$4i$ (correct answer)
$-4$
$4$

a^2

+ ab +

b^2

) holds over complexes, with identical algebra. For a=1+i, b=1-i, a-b=2i, a^2 + ab + b^2=2, so 2i * 2=4i, verified by direct expansion. Choice B correctly computes 4i as the value. A tempting distractor like A might negate incorrectly, perhaps mishandling signs in a-b or i powers. Transferable strategy: Verify cube differences: (1) Find a-b. (2) Compute a^2, ab, b^2 separately with i^2=-1. (3) Add and multiply. (4) Check against direct (

a^3

b^3

). Keep up the excellent progress!

Question 6

A student claims the identity $a^2-b^2=(a+b)(a-b)$ works for complex numbers as well. Using $a=1+i$ and $b=2-i$ , which option gives the correct value of $a^2-b^2$ (and matches the product $(a+b)(a-b)$ )?

$-3+6i$ (correct answer)
$-3-6i$
$3+6i$
$3-6i$

Explanation: This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, allowing us to compute differences like a^2 - b^2 with complex values. All polynomial identities that work for real numbers also work for complex numbers because complex numbers form an algebraically closed field—addition, subtraction, multiplication work with same properties (commutative, associative, distributive). For example, with a=1+i and b=2-i, a^2=2i and b^2=3-4i, so a^2 - b^2=-3+6i, which matches (a+b)(a-b)=3(-1+2i)=-3+6i. Choice A correctly applies the polynomial identity and computes a^2 - b^2 as -3+6i, verifying it equals the factored form over complex numbers. Choice C might tempt if you flip signs in subtraction, but remember to distribute the negative carefully when computing a^2 - b^2—track real and imaginary parts separately. Verifying polynomial identities with complex values: (1) Choose specific complex values; (2) Calculate left side directly; (3) Calculate right side using identity; (4) Simplify both with i^2=-1; (5) Compare for equality. Great job exploring this—you're seeing how complex numbers make identities even more powerful!

Question 7

Use the difference of squares identity to factor $x^2+4$ completely over $\mathbb{C}$ (even though it does not factor over the reals).

$x^2+4=(x+2)(x-2)$
$x^2+4=(x+4i)(x-4i)$
$x^2+4=(x+2i)(x-2i)$ (correct answer)
$x^2+4=(x+i)^2$

x^2 + 4

doesn't factor over reals (sum of squares), but over complex numbers we can use difference of squares with

i

x^2 + 4 = x^2 - (-4) = x^2 - (2i)^2

because

(2i)^2 = 4i^2 = -4

. Now apply

a^2 - b^2 = (a+b)(a-b)

with

a = x

b = 2i

x^2 + 4 = (x + 2i)(x - 2i)

. Verify:

(x + 2i)(x - 2i) = x^2 - 2xi + 2xi - 4i^2 = x^2 - 4(-1) = x^2 + 4

. Perfect! Choice C correctly identifies this factorization. Choice B makes an error with

(x+4i)(x-4i)

, which would give

x^2 - 16i^2 = x^2 + 16

, not

x^2 + 4

—when factoring

x^2 + k

, you need factors

(x + \sqrt{k}i)(x - \sqrt{k}i)

, not

(x + ki)(x - ki)

! Factoring sums of squares over complex: (1) Recognize

a^2 + b^2

as target. (2) Find

\sqrt{b}

to write as

(\sqrt{b}i)^2 = -b

. (3) Apply difference of squares. The beauty: every sum of squares factors over complex numbers using conjugate pairs!

Question 8

Factor completely over the complex numbers: $x^4+4$ . (Hint: use $a^2+b^2=(a+bi)(a-bi)$ with $a=x^2$ and $b=2$ .)

$(x^2+2)(x^2-2)$
$(x^2+2i)(x^2-2i)$ (correct answer)
$(x+2i)(x-2i)(x+2)(x-2)$
$(x^2+4i)(x^2-4i)$

x^2

+2i)(

x^2

-2i), expanding to x^4 +4. Choice C attempts linear factors but multiplies to x^4 -16—check expansions carefully! For complete factoring over complex: (1) Recognize as sum of squares with a=x^2; (2) Use (a + bi)(a - bi); (3) If needed, factor quadratics further using roots; (4) Verify product. You're amazing— this shows how identities help factor polynomials fully over complex numbers!

Question 9

Which option correctly applies the identity $(a+b)^2 = a^2 + 2ab + b^2$ with complex numbers $a = i$ and $b = 2 - i$ to compute $(i + (2 - i))^2$ ?

$4$ (correct answer)
$-4$
$4i$
$-4i$

(a+b)^2

with fully complex a and b. All polynomial identities that work for real numbers also work for complex numbers because complex numbers form an algebraically closed field—addition, subtraction, multiplication work with same properties (commutative, associative, distributive). With

a=i

b=2-i

, direct sum is

2

, squared is

4

; via identity,

a^2 + 2ab + b^2 = -1 + (2 + 4i) + (3 - 4i) = 4

. Choice A correctly applies the polynomial identity to get

4

, with imaginary parts canceling perfectly. Choice C might come from mishandling signs in

2ab

—compute products step by step, grouping reals and imaginaries. Verifying with complex values: (1) Simplify left side directly; (2) Expand right side term by term; (3) Apply

i^2 = -1

; (4) Add and compare. You're fantastic—seeing these matches reinforces the power of extending identities to complex numbers!

Question 10

The identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ works for complex $a, b$ as well. Let $a = i$ and $b = 1$ . What is the value of $a^3 + b^3$ ?

$1 + i$
$1 - i$ (correct answer)
$-1 - i$
$-1 + i$

a^3 + b^3

a = i

and

b = 1

a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (1 + i)(-i) = 1 - i

, matching direct computation:

(-i) + 1 = 1 - i

. Choice A correctly applies the polynomial identity to yield

1 - i

through precise i powers. A tempting distractor like choice B=

1 + i

might flip the sign by miscounting

i^3

as i instead of

-i

; remember the cycle:

i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

. To apply sum of cubes with complexes: (1) Compute

a + b

. (2) Find

a^2

-ab

b^2

and sum. (3) Multiply and simplify. Wonderful job—these examples prove identities are universal!

Question 11

The identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ works for complex $a,b$ as well. Let $a=i$ and $b=1$ . What is the value of $a^3+b^3$ ?

$1-i$ (correct answer)
$1+i$
$-1+i$
$-1-i$

a^2

- ab +

b^2

) = (1+i)(-i) =1-i, matching direct (-i)+1=1-i. Choice A correctly applies the polynomial identity to yield 1-i through precise i powers. A tempting distractor like choice B=1+i might flip the sign by miscounting i^3 as i instead of -i; remember the cycle: i^1=i, i^2=-1, i^3=-i, i^4=1. To apply sum of cubes with complexes: (1) Compute a+b. (2) Find a^2, -ab, b^2 and sum. (3) Multiply and simplify. Wonderful job—these examples prove identities are universal!

Question 12

Use the identity $a^2-b^2=(a+b)(a-b)$ with complex numbers to compute $(1+i)^2-(2-i)^2$ by factoring. Which result is correct?

$-3-6i$
$3+6i$
$3-6i$
$-3+6i$ (correct answer)

Explanation: This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, enabling computation via factoring. Key concept: The difference of squares a^2 - b^2 = (a+b)(a-b) works for complex a and b, as operations are consistent. Here, with a=1+i, b=2-i, (a+b)(a-b)=(3)(-1+2i)=-3+6i, matching direct computation. Choice A correctly gives -3+6i as the result. A tempting distractor like B might flip signs, forgetting to carefully compute a-b as (1+i)-(2-i)=-1+2i. Transferable strategy: For complex differences: (1) Compute a+b and a-b separately, grouping real and imaginary parts. (2) Multiply using distributive property, simplify i^2=-1. (3) Combine terms. You're building strong skills in complex arithmetic!

Question 13

A key idea is that identities like $a^2+b^2=(a+bi)(a-bi)$ work in $\mathbb{C}$ . Factor $x^2+16$ completely over $\mathbb{C}$ .

$x^2+16=(x+4i)^2$
$x^2+16=(x+8i)(x-8i)$
$x^2+16=(x+4i)(x-4i)$ (correct answer)
$x^2+16=(x+4)(x-4)$

x^2 + 16

doesn't factor over reals (sum of squares), but over complex numbers we can use difference of squares with

i

x^2 + 16 = x^2 - (-16) = x^2 - (4i)^2

because

(4i)^2 = 16i^2 = -16

. Now apply

a^2 - b^2 = (a+b)(a-b)

with

a = x

b = 4i

x^2 + 16 = (x + 4i)(x - 4i)

. Verify:

(x + 4i)(x - 4i) = x^2 - 4xi + 4xi - 16i^2 = x^2 - 16(-1) = x^2 + 16

. Perfect! Choice C correctly applies this factorization. Choice B incorrectly uses

(x+8i)(x-8i)

, which would give

x^2 - 64i^2 = x^2 + 64

, not

x^2 + 16

—when factoring

x^2 + k

, the factors are

(x + \sqrt{k}i)(x - \sqrt{k}i)

, so you need

\sqrt{16} = 4

, not

16/2 = 8

. Factoring sums of squares over complex: (1) Recognize

a^2 + b^2

as target. (2) Rewrite as

a^2 - (bi)^2

because

(bi)^2 = -b^2

. (3) Apply difference of squares:

(a + bi)(a - bi)

. Every sum of squares factors into conjugate linear factors over complex!

Question 14

Factor completely over $\mathbb{C}$ using standard identities. $x^4+4$ (Hint: rewrite as a difference of squares: $x^4+4=(x^2+2)^2-(2x)^2$ .)

$(x^2+2x+2)(x^2-2x+2)$ (correct answer)
$(x^2+2)(x^2+2)$
$(x^2-2x-2)(x^2+2x-2)$
$(x^2+2x-2)(x^2-2x-2)$

x^4 + 4

as a difference of squares. Following the hint:

x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2

. Now apply difference of squares

a^2 - b^2 = (a+b)(a-b)

with

a = x^2 + 2

and

b = 2x

x^4 + 4 = ((x^2+2) + 2x)((x^2+2) - 2x) = (x^2 + 2x + 2)(x^2 - 2x + 2)

. Verify by expanding:

(x^2+2x+2)(x^2-2x+2) = x^4 - 2x^3 + 2x^2 + 2x^3 - 4x^2 + 4x + 2x^2 - 4x + 4 = x^4 + 4

. Perfect! Choice A correctly shows the factorization as

(x^2+2x+2)(x^2-2x+2)

. Choice B incorrectly suggests

(x^2+2)(x^2+2)

, but this would give

x^4 + 4x^2 + 4

, not

x^4 + 4

. Choices C and D have wrong signs that don't produce the correct expansion. Advanced factoring technique: (1) Add and subtract a strategic middle term to create a perfect square. (2) Rewrite as difference of squares. (3) Apply the difference of squares identity. (4) Each resulting quadratic can be further factored over complex into linear factors! This Sophie Germain identity shows how creative algebraic manipulation extends our factoring power!

Question 15

Using the identity $a^2+b^2=(a+bi)(a-bi)$ (valid over complex numbers), which option correctly factors $x^2+16$ over $\mathbb{C}$ ?

$(x+4)(x-4)$
$(x+8i)(x-2i)$
$(x+4i)(x-4i)$ (correct answer)
$(x+2i)(x-2i)$

Explanation: This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, allowing us to factor sums of squares using a^2 + b^2 = (a+bi)(a-bi). All polynomial identities that work for real numbers also work for complex numbers because complex numbers form an algebraically closed field—addition, subtraction, multiplication work with same properties (commutative, associative, distributive). For x^2 +16, it's x^2 + (4)^2 = (x + 4i)(x - 4i), since (4i)^2 =16i^2=-16, so x^2 - (-16). Choice C correctly applies the polynomial identity to factor as (x+4i)(x-4i), verifying by expansion to x^2 +16. Choice D uses 2 instead of 4, but match the coefficient: b^2=16 means b=4 for the identity. Applying the sum of squares identity: (1) Set a=x, b= constant; (2) Form (a + bi)(a - bi); (3) Expand to confirm; (4) Note the conjugate pairs. You're doing wonderfully— this technique opens up factoring for many quadratics!

Question 16

Over $\mathbb{C}$ , the sum of squares factors using $a^2+b^2=(a+bi)(a-bi)$ . Using this idea, which is a correct factorization of $x^2+16$ over the complex numbers?

$(x+4i)(x-4i)$ (correct answer)
$(x+4)(x-4)$
$(x+4i)^2$
$(x+8i)(x-8i)$

Explanation: This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, factoring sums via imaginaries. Key concept: a^2 + b^2 = (a + bi)(a - bi), since (bi)^2 = -b^2. For x^2 +16 = x^2 + (4)^2, use (x + 4i)(x - 4i), as (4i)^2=-16. Choice C correctly applies this pattern. A tempting distractor like B uses 8i instead of 4i, mismatching (8i)^2=-64 for +16—scale correctly! Transferable strategy: Identify b in +b^2, set to (sqrt(b) i)^2 = -b^2, factor difference. Fantastic effort expanding your toolkit!

Question 17

Consider the polynomial $P(x) = x^6 - 64$ . When factored completely over the complex numbers, how many distinct linear factors does $P(x)$ have?

4
6 (correct answer)
3
2

Explanation: The polynomial

P(x) = x^6 - 64 = x^6 - 2^6

can be factored as a difference of sixth powers. First,

x^6 - 64 = (x^3)^2 - 8^2 = (x^3 + 8)(x^3 - 8)

. Then

x^3 + 8 = (x + 2)(x^2 - 2x + 4)

and

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

. The quadratic factors

x^2 - 2x + 4

and

x^2 + 2x + 4

each have two complex roots (since their discriminants are negative). Therefore, we have 6 distinct linear factors total when factored completely over the complex numbers. Choice A counts only the real factors plus some complex factors. Choice C incorrectly assumes some factors are repeated. Choice D only counts the obvious real factors.

Question 18

If $f(x) = x^3 + 8$ and $g(x) = x + 2$ , then $\frac{f(x)}{g(x)}$ can be written as a polynomial for all values of $x$ except $x = -2$ . What is this polynomial?

$x^2 - 2x + 4$ (correct answer)
$x^2 + 2x + 4$
$x^2 - 2x - 4$
$x^2 + 4x + 8$

Explanation: Since

f(x) = x^3 + 8

, we recognize this as a sum of cubes:

x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)

. Therefore,

\frac{f(x)}{g(x)} = \frac{x^3 + 8}{x + 2} = \frac{(x + 2)(x^2 - 2x + 4)}{x + 2} = x^2 - 2x + 4

for

x \neq -2

. Choice B incorrectly uses the sum of cubes formula with wrong signs. Choice C has an incorrect constant term. Choice D results from incorrectly expanding

(x + 2)^2 + 4

Question 19

Which of the following represents the extension of the polynomial identity $(a^2 + b^2) = (a + bi)(a - bi)$ when applied to the expression $x^4 + x^2 + 1$ ?

$(x^2 + \frac{1 + \sqrt{3}i}{2}x + 1)(x^2 + \frac{1 - \sqrt{3}i}{2}x + 1)$
$(x^2 + xi + 1)(x^2 - xi + 1)$
$(x^2 + x + 1)(x^2 - x + 1)$ (correct answer)
$(x^2 + 1 + xi)(x^2 + 1 - xi)$

Explanation: When you encounter polynomial factoring problems involving complex expressions, look for ways to apply familiar identities like the sum of squares formula

(a^2 + b^2) = (a + bi)(a - bi)

, but also consider simpler real factorizations first. To factor

x^4 + x^2 + 1

, let's multiply it by

(x^2 - 1)

and see what happens:

(x^4 + x^2 + 1)(x^2 - 1) = x^6 - 1

Since

x^6 - 1 = (x^2)^3 - 1^3 = (x^2 - 1)(x^4 + x^2 + 1)

, we can use the difference of cubes formula:

x^6 - 1 = (x^3 - 1)(x^3 + 1)

. Further factoring:

x^6 - 1 = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)

Since

(x^2 - 1) = (x - 1)(x + 1)

, we can divide both sides by

(x^2 - 1)

to get:

x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)

Let's verify by expanding:

(x^2 + x + 1)(x^2 - x + 1) = x^4 - x^3 + x^2 + x^3 - x^2 + x + x^2 - x + 1 = x^4 + x^2 + 1

✓ Answer A uses complex coefficients unnecessarily when a real factorization exists. Answer B incorrectly places

xi

in the middle terms rather than treating this as a quadratic-type expression. Answer D attempts to factor as a sum of squares but uses the wrong structure entirely. Strategy tip: When factoring quartic polynomials, try relating them to differences of cubes or sixth powers first—this often reveals simpler real factorizations before resorting to complex methods.

Question 20

The polynomial identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ can be extended to complex numbers. Using this identity, what is the complete factorization of $x^3 - 27i$ over the complex numbers?

$(x - 3i^{1/3})(x^2 + 3i^{1/3}x + 9i^{2/3})$
$(x - 3i)(x^2 + 3ix + 9i^2)$
$(x + 3i)(x^2 - 3ix + 9i^2)$
$(x - 3\sqrt[3]{i})(x^2 + 3x\sqrt[3]{i} + 9\sqrt[3]{i^2})$ (correct answer)

Explanation: To use the identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

, we need to identify

a = x

and

b^3 = 27i

. Since

27i = 27 \cdot i

, we need

b = \sqrt[3]{27i} = 3\sqrt[3]{i}

. Applying the identity:

x^3 - 27i = (x - 3\sqrt[3]{i})(x^2 + x \cdot 3\sqrt[3]{i} + (3\sqrt[3]{i})^2) = (x - 3\sqrt[3]{i})(x^2 + 3x\sqrt[3]{i} + 9\sqrt[3]{i^2})

. Choice A uses incorrect notation for cube roots. Choice B incorrectly assumes

\sqrt[3]{27i} = 3i

. Choice C has wrong signs in the factorization.