Algebra 2 Quiz: Extending Polynomial Identities To Complex Numbers

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Extending Polynomial Identities To Complex NumbersQuestion 1 of 20

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

-5-12i

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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

$13-12i$
$13+12i$
$-5+12i$ (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 2

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 3

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 4

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 5

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 6

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 7

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 8

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.

Question 9

The polynomial $f(x) = x^8 - 256$ can be factored completely over the complex numbers into linear factors of the form $(x - r)$ where $r$ represents the 8th roots of 256. How many of these roots are purely imaginary (have zero real part)?

$0$
$4$
$2$ (correct answer)
$8$

Explanation: When you encounter a polynomial like

f(x) = x^8 - 256

, you're dealing with finding complex roots using De Moivre's theorem and the unit circle. The equation

x^8 = 256

has 8 solutions that are evenly spaced around a circle in the complex plane. First, express 256 in polar form:

256 = 256e^{i \cdot 0}

. The 8th roots are found using

x_k = \sqrt[8]{256} \cdot e^{i(0 + 2\pi k)/8}

where

k = 0, 1, 2, ..., 7

. Since

\sqrt[8]{256} = 2

, the roots are

x_k = 2e^{i\pi k/4}

. These eight roots correspond to angles:

0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}

. Converting to rectangular form, purely imaginary numbers have zero real part, which occurs when

\cos(\theta) = 0

. This happens at

\theta = \frac{\pi}{2}

and

\theta = \frac{3\pi}{2}

, giving us the roots

2i

and

-2i

. Answer choice A) 0 incorrectly assumes no imaginary roots exist. Answer choice B) 4 mistakenly counts all non-real roots (including those with both real and imaginary parts). Answer choice D) 8 confuses the total number of roots with purely imaginary ones. The correct answer is C) 2. Study tip: For

n

th roots of real numbers, purely imaginary roots occur only when the angle puts you at

\frac{\pi}{2}

\frac{3\pi}{2}

on the unit circle—look for where cosine equals zero.

Question 10

If $z^2 + 4z + 13 = 0$ , then $z$ can be written in the form $a + bi$ where $a$ and $b$ are real numbers. The expression $z^2 + 4z + 13$ can be factored as $(z - (a + bi))(z - (a - bi))$ . What are the values of $a$ and $b$ ?

$a = 4, b = 13$
$a = 2, b = 3$
$a = -2, b = -3$
$a = -2, b = 3$ (correct answer)

Explanation: When you encounter a quadratic equation that doesn't factor nicely with real numbers, you're dealing with complex solutions. The key insight here is using the quadratic formula and understanding how complex conjugates work as roots. To find the values of

a

and

b

, use the quadratic formula on

z^2 + 4z + 13 = 0

z = \frac{-4 \pm \sqrt{16 - 4(1)(13)}}{2} = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2}

Since

\sqrt{-36} = 6i

, we get:

z = \frac{-4 \pm 6i}{2} = -2 \pm 3i

So the two solutions are

z = -2 + 3i

and

z = -2 - 3i

. These are complex conjugates in the form

a + bi

and

a - bi

, where

a = -2

and

b = 3

. Choice A)

a = 4, b = 13

incorrectly uses the coefficients from the original equation without applying the quadratic formula. Choice B)

a = 2, b = 3

gets the magnitude of

a

right but misses the negative sign from

-4/2

. Choice C)

a = -2, b = -3

correctly finds

a

but makes

b

negative—remember that

b

represents the coefficient of the imaginary unit in the standard form, so it's positive even though one root has

-3i

. The answer is D)

a = -2, b = 3

. Study tip: When solving quadratic equations with complex roots, always write your final answer as

a \pm bi

to clearly identify the real part

a

and imaginary coefficient

b

Question 11

Over the complex numbers, the difference of squares identity $a^2-b^2=(a+b)(a-b)$ still applies. Which option correctly factors $x^2+9$ over $\mathbb{C}$ ?

$(x+9)(x-1)$
$x^2+9$ does not factor over the complex numbers.
$(x+3i)(x-3i)$ (correct answer)
$(x+3)(x-3)$

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. 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). The power of complex extension: x^2 +9 doesn't factor over reals (sum of squares), but over complex numbers we can use difference of squares with i: x^2 +9 = x^2 - (-9) = x^2 - (3i)^2, since (3i)^2 =9i^2=-9. Choice C correctly applies the polynomial identity to factor over complex numbers as (x+3i)(x-3i), which expands back to x^2 +9 using i^2=-1. Choice B claims it doesn't factor, forgetting that over complex numbers every sum of squares factors using the extended identity—don't limit yourself to real factors! Factoring sums of squares over complex: (1) Recognize a^2 + b^2 as target; (2) Rewrite as a^2 - (bi)^2; (3) Apply difference of squares: (a + bi)(a - bi); (4) Verify by expanding. Keep practicing these— you're unlocking new ways to factor and solve equations with complex numbers!

Question 12

A polynomial identity such as $(a+b)^2=a^2+2ab+b^2$ is known to hold for real numbers. Which option correctly verifies that it also holds for complex numbers by using $a=1+i$ and $b=2-i$ ?

LHS: $(1+i+2-i)^2=3^2=9$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=2i+(6-2i)+(3-4i)=9-4i$ .
LHS: $(1+i+2-i)^2=3^2=9$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=2i+(6+2i)+(3-4i)=9$ . (correct answer)
The identity only holds for real $a,b$ , so substituting $a=1+i$ and $b=2-i$ is invalid.
LHS: $(1+i+2-i)^2=3^2=9$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=2i+(6+2i)+(3+4i)=9+8i$ .

a=1+i

and

b=2-i

, the left side simplifies to

(3)^2=9

, and the right side requires careful computation of each term, remembering

i^2=-1

. Choice A correctly applies the polynomial identity by accurately calculating each part of the right side—

(1+i)^2=2i

2(1+i)(2-i)=6+2i

(2-i)^2=3-4i

—and summing to 9, matching the left side. Choice B makes a sign error in

(2-i)^2

, using

+4i

instead of

-4i

, which throws off the imaginary parts and prevents cancellation. To verify identities with complex numbers, always compute both sides independently: simplify the left side directly, expand the right side term by term, and ensure

i^2

is replaced with

-1

immediately to avoid mistakes. You're doing great—practicing with specific values like these builds confidence in how seamlessly algebra extends to complex numbers!

Question 13

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 14

Polynomial identities that hold for real numbers also hold for complex numbers because addition and multiplication work the same way in $\mathbb{C}$ . Use the identity $(a+b)^2=a^2+2ab+b^2$ to verify it for the complex numbers $a=1+i$ and $b=2-i$ . Which option correctly shows that both sides simplify to the same complex number?

LHS: $(a+b)^2=(3)^2=9$ . RHS: $a^2+2ab+b^2=(1+i)^2+2(1+i)(2-i)+(2-i)^2=9$ .
LHS: $(a+b)^2=(3)^2=9$ . RHS: $a^2+2ab+b^2=(1+i)^2+2(1+i)(2-i)+(2-i)^2=9+2i$ .
LHS: $(a+b)^2=(1+i+2-i)^2=3^2=9$ . RHS: $a^2+2ab+b^2=2i+2(3+i)+(3-4i)=9$ . (correct answer)
LHS: $(a+b)^2=(3)^2=9$ . RHS: $a^2+2ab+b^2=(1+i)^2+2(1+i)(2-i)+(2-i)^2=7$ .

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

with

a = 1+i

and

b = 2-i

: Left side:

(a+b)^2 = ((1+i)+(2-i))^2 = (3)^2 = 9

. Right side:

a^2 + 2ab + b^2 = (1+i)^2 + 2(1+i)(2-i) + (2-i)^2 = (1+2i-1) + 2(2-i+2i+1) + (4-4i-1) = 2i + 2(3+i) + (3-4i) = 2i + 6 + 2i + 3 - 4i = 9

. Perfect match! Choice C correctly shows this calculation, properly expanding

(1+i)^2 = 1+2i+i^2 = 2i

and tracking all terms. Choice B makes an error by getting

9+2i

on the right side, forgetting to combine all real and imaginary parts—when verifying identities with complex numbers, always group real parts together and imaginary parts together before comparing. The key insight: polynomial identities are universal truths about algebraic operations, not limited to any particular number system. When extending to complex numbers, use the same algebraic rules but remember

i^2 = -1

throughout your calculations!

Question 15

Polynomial identities that are true for real numbers also hold for complex numbers because addition and multiplication work the same way in $\mathbb{C}$ . Use the identity $(a+b)^2=a^2+2ab+b^2$ to verify it for the complex values $a=1+i$ and $b=2-i$ . Which option shows a correct verification (both sides match)?

LHS: $(a+b)^2=(3)^2=9$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=9$ . (correct answer)
LHS: $(a+b)^2=(3)^2=9$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=7$ .
LHS: $(a+b)^2=(3)^2=9$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=9+2i$ .
LHS: $(a+b)^2=(3)^2=9i$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=9i$ .

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

with

a = 1+i

and

b = 2-i

: Left side:

(a+b)^2 = ((1+i)+(2-i))^2 = (3)^2 = 9

. Right side:

(1+i)^2 + 2(1+i)(2-i) + (2-i)^2 = (1+2i+i^2) + 2(2-i+2i-i^2) + (4-4i+i^2) = (1+2i-1) + 2(2-i+2i+1) + (4-4i-1) = 2i + 2(3+i) + (3-4i) = 2i + 6 + 2i + 3 - 4i = 9

. Perfect match! Choice A correctly shows both sides equal 9, confirming the identity extends seamlessly to complex numbers. Choice B makes a calculation error on the right side, getting 7 instead of 9—likely missing terms when expanding. Choice C incorrectly leaves an imaginary part in the final answer, forgetting that when we add conjugates like

(1+i)

and

(2-i)

, imaginary parts can cancel. The beauty of polynomial identities: they're universal truths that transcend number systems, working equally well whether your variables are integers, rationals, reals, or complex numbers!

Question 16

Use the difference of cubes identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ with complex numbers. Let $a=1+i$ and $b=1-i$ . What is $a^3-b^3$ ?

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

a^3 - b^3

with

a = 1+i

and

b = 1-i

, we can use the difference of cubes identity

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

, or compute directly. From the previous problem, we found

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

and

b^3 = (1-i)^3 = -2 - 2i

. Therefore,

a^3 - b^3 = (-2+2i) - (-2-2i) = -2 + 2i + 2 + 2i = 0 + 4i = 4i

. Perfect! Choice C correctly identifies

4i

as the answer. Choice D would give

-4i

, likely from a sign error when subtracting—remember that subtracting a negative imaginary part makes it positive:

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

. The difference of cubes for complex conjugates: when

a

and

b

are conjugates like

1+i

and

1-i

, their cubes are also conjugates, so

a^3 - b^3

is purely imaginary (real parts cancel, imaginary parts double). This pattern extends: for conjugates, odd power differences are purely imaginary, even power differences are purely real!

Question 17

The expression $9x^2 + 36$ can be factored over the complex numbers as $9(x + ai)(x - ai)$ where $a$ is a positive real number. What is the value of $a$ ?

$2$ (correct answer)
$6$
$3$
$4$

Explanation: First, factor out the common factor:

9x^2 + 36 = 9(x^2 + 4)

. To factor

x^2 + 4

over the complex numbers, we need

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

. Therefore,

9x^2 + 36 = 9(x + 2i)(x - 2i)

, which matches the form

9(x + ai)(x - ai)

with

a = 2

. Choice B results from confusing the coefficient 36 with the value under the square root. Choice C comes from taking the square root of 9. Choice D results from taking the square root of 16 instead of 4.

Question 18

Use polynomial identities over complex numbers to factor $x^4+4$ completely over $\mathbb{C}$ . (Hint: $x^4+4=(x^2)^2- (2i)^2$ .) Which option gives a correct complete factorization into linear factors?

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

x^2

-2i)(

x^2

+2i), then factor each quadratic over C. The linear factors are roots like ±(1+i), ±(1-i), matching the expansion. Choice D correctly provides this complete factorization into linears. A tempting distractor like B uses √2 and i√2, leading to x^4 -4 instead—verify constants! Transferable strategy: For x^4 + k, add/subtract terms or use difference of squares with i, then factor quadratics. Amazing job tackling higher degrees!

Question 19

A quadratic with no real zeros can still factor over $\mathbb{C}$ . Which option correctly factors $P(x)=x^2+2x+5$ over the complex numbers?

$(x+1-2i)(x+1+2i)$ (correct answer)
$(x-1-2i)(x-1+2i)$
$(x+1-4i)(x+1+4i)$
$x^2+2x+5$ does not factor over $\mathbb{C}$ .

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 factoring of quadratics with negative discriminants over complex. 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 +2x +5, complete the square or use quadratic formula: roots -1 ± 2i, leading to linear factors. Choice A correctly applies the polynomial identity principles to factor as (x+1-2i)(x+1+2i), expanding to x^2 +2x +5. Choice D claims it doesn't factor, but over complex, every polynomial factors completely into linears—use the formula! Factoring quadratics over complex: (1) Compute discriminant; (2) If negative, express as d = k i^2 with k>0; (3) Roots [-b ± sqrt(|d|) i]/2a; (4) Write as (x - root1)(x - root2). Excellent job—you're discovering how complex numbers ensure all quadratics factor!

Question 20

Over the complex numbers, the difference of squares identity $a^2-b^2=(a+b)(a-b)$ still holds. Factor the polynomial $x^2+9$ completely over $\mathbb{C}$ .

It does not factor over $\mathbb{C}$ .
$(x+3i)(x-3i)$ (correct answer)
$(x+3)(x-3)$
$(x+9)(x-1)$

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. 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). The power of complex extension: x^2 +9 doesn't factor over reals (sum of squares), but over complex numbers we can use difference of squares with i: x^2 +9 = x^2 - (-9) = x^2 - (3i)^2, then (x + 3i)(x - 3i). Choice C correctly applies the polynomial identity to factor over complex numbers using the difference of squares extended to the complex domain. A tempting distractor like choice B might claim (x+3)(x-3), but that's for x^2-9; remember, over reals, x^2+9 has no factors, but complexes unlock it. Factoring sums of squares over complex: (1) Recognize a^2 + b^2 as target. (2) Rewrite as a^2 - (bi)^2. (3) Apply difference of squares: (a + bi)(a - bi). You're doing amazing—keep practicing to see how complexes make polynomials factor completely!