Algebra 2 Quiz: Fundamental Theorem Of Algebra For Quadratics

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Fundamental Theorem Of Algebra For QuadraticsQuestion 1 of 20

A quadratic equation has the form $ax^2+bx+c=0$ with real coefficients. In $\mathbb{C}$ , every quadratic has exactly 2 solutions counting multiplicity (two distinct real, one repeated real with multiplicity 2, or two complex conjugates). For $x^2+4x+4=0$ , use the discriminant $b^2-4ac$ to predict the solution type and state the solutions counting multiplicity.

Which statement is correct?

Discriminant

>0

, so there are two distinct real solutions:

x=-2

and

x=2

Discriminant

=0

, so there is one repeated real solution:

x=-2

with multiplicity 2 (total of 2 solutions counted with multiplicity)

Discriminant

<0

, so there are two complex conjugate solutions:

x=-2\pm 2i

There is exactly 1 solution in

\mathbb{C}

because the quadratic is a perfect square

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Algebra 2 Quiz

Algebra 2 Quiz: Fundamental Theorem Of Algebra For Quadratics

Practice Fundamental Theorem Of Algebra For Quadratics 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 Fundamental Theorem Of Algebra For Quadratics, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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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

Which statement is correct?

Discriminant $>0$ , so there are two distinct real solutions: $x=-2$ and $x=2$
Discriminant $=0$ , so there is one repeated real solution: $x=-2$ with multiplicity 2 (total of 2 solutions counted with multiplicity) (correct answer)
Discriminant $<0$ , so there are two complex conjugate solutions: $x=-2\pm 2i$
There is exactly 1 solution in $\mathbb{C}$ because the quadratic is a perfect square

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2 (like x = 3 counted twice from (x - 3)²), (3) if negative, two complex conjugate solutions. For x² + 4x + 4 = 0, let's verify exactly 2 complex solutions: (1) Calculate discriminant: b² - 4ac = 16 - 16 = 0 (zero). (2) This means 1 real solution with multiplicity 2. (3) Notice x² + 4x + 4 = (x + 2)² = 0, giving x = -2 with multiplicity 2. (4) Count: 2 solutions (one value appearing twice). Choice B correctly recognizes that a zero discriminant means one repeated real solution with multiplicity 2, giving the total count of 2 required by the Fundamental Theorem. Choice A incorrectly claims positive discriminant and two distinct solutions—but (x + 2)² = 0 clearly gives only x = -2 repeated! Choice C incorrectly claims negative discriminant and complex solutions when the perfect square factorization shows real solutions. Choice D makes the critical error of counting a repeated root as only 1 solution total, violating the Fundamental Theorem which requires counting multiplicity. Solution counting with multiplicity: (1) When discriminant = 0, the quadratic is a perfect square. (2) Factor as a(x - r)² = 0, giving solution r with multiplicity 2. (3) Count that value twice: one distinct value, but 2 total solutions. (4) This maintains the requirement of exactly 2 solutions for degree 2. Perfect square recognition: ax² + bx + c with discriminant 0 can be written as a(x - h)² where h = -b/(2a). The repeated root appears twice in the factorization, so we count it twice!

Question 2

Factor the quadratic $x^2+25$ over the complex numbers. Recall: every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity (either 2 real, 1 repeated real counted twice, or 2 complex conjugates).

Which option gives the correct factorization over $\mathbb{C}$ and the zeros?

$x^2+25=(x+5)(x-5)$ ; zeros: $5,-5$
$x^2+25=(x+5i)(x-5i)$ ; zeros: $5i,-5i$ (correct answer)
$x^2+25=(x+25i)(x-25i)$ ; zeros: $25i,-25i$
It cannot be factored over $\mathbb{C}$ because it has no real roots

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—even when there are no real solutions! To factor x² + 25, we need to find its zeros by solving x² + 25 = 0, which gives x² = -25. Taking square roots: x = ±√-25 = ±5i (since √-25 = √25 · √-1 = 5i). Therefore, x² + 25 = (x - 5i)(x - (-5i)) = (x - 5i)(x + 5i), with zeros at 5i and -5i. Choice B correctly identifies the factorization (x + 5i)(x - 5i) with zeros 5i and -5i—note that these are complex conjugates, as required for a real-coefficient quadratic. Choice A incorrectly factors as (x + 5)(x - 5) = x² - 25, not x² + 25—this is the difference of squares formula used backwards! Choice C has the wrong zeros (25i instead of 5i), arising from the error √25 = 25. Choice D incorrectly claims no factorization is possible over ℂ, but the Fundamental Theorem guarantees every polynomial factors completely over the complex numbers! Factoring with complex zeros: (1) Set the quadratic equal to zero and solve for x. (2) For x² + k² = 0, the zeros are ±ki (pure imaginary). (3) Write as product of linear factors: (x - zero₁)(x - zero₂). (4) Verify by expanding: (x - ki)(x + ki) = x² - (ki)² = x² - k²i² = x² + k². Sum and difference patterns: x² + k² = (x + ki)(x - ki) (sum of squares factors over ℂ), while x² - k² = (x + k)(x - k) (difference of squares factors over ℝ). The key difference is that sum of squares requires complex factors!

Question 3

Which example correctly represents the case of a quadratic having exactly 2 solutions in the complex numbers but only 1 distinct solution value because it has multiplicity 2? (Recall: for $ax^2+bx+c=0$ , the discriminant $b^2-4ac$ determines whether there are 2 distinct real solutions, 1 repeated real solution counted twice, or 2 complex conjugates.)

$x^2-5x+6=0$ has solutions $2$ and $3$
$x^2+4=0$ has solutions $2i$ and $-2i$
$(x-3)^2=0$ has solution $x=3$ with multiplicity 2 (correct answer)
$x^2+1=0$ has one solution $i$ (multiplicity 1)

Question 4

A quadratic with real coefficients has the form $ax^2+bx+c=0$ and always has exactly 2 complex solutions counting multiplicity (2 distinct real, 1 repeated real, or 2 complex conjugates). Suppose a real-coefficient quadratic has one solution $2+3i$ . What must the other solution be, and why?

$2-3i$ , because non-real solutions of real-coefficient quadratics occur in complex conjugate pairs. (correct answer)
$-2-3i$ , because you negate both parts to get the other root.
$2+3i$ again, because complex roots cannot repeat.
There is no second solution, because a quadratic can have only one complex root.

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. For a real-coefficient quadratic with one solution 2 + 3i, the other must be 2 - 3i, as non-real roots come in conjugate pairs to ensure real coefficients. Choice A correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the conjugate pair requirement for real coefficients. Choice D says no second solution because a quadratic can have only one complex root—this fails because the theorem guarantees exactly 2, and for real coefficients, complex roots pair up as conjugates; you can't have just one. To apply this transferable strategy: (1) If given one complex root for real-coefficient quadratic, the other is its conjugate (flip imaginary sign). (2) Verify by plugging in or using Vieta's formulas. For example, if root is 4 + i, other is 4 - i. You're building strong skills—keep going!

Question 5

Use the discriminant to determine the solution type for $x^2-5x+6=0$ . Remember: every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity (2 distinct real, 1 repeated real, or 2 complex conjugates), and the discriminant $b^2-4ac$ determines which case. Which statement is correct?

The discriminant is negative, so the solutions are a complex conjugate pair.
The discriminant is $0$ , so there is one repeated real solution.
The discriminant is positive, so there are two distinct real solutions (2 total solutions). (correct answer)
Since it is a quadratic, it could have 0, 1, or 2 complex solutions depending on the discriminant.

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. For x squared - 5x + 6 = 0, discriminant = 25 - 24 = 1 (positive), so two distinct real solutions (x=2, x=3), totaling 2. Choice C correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the distinct real case from positive discriminant. Choice D says it could have 0, 1, or 2 complex solutions depending on discriminant—this is misleading; over complexes, always exactly 2 total (which may be real or complex). To apply this transferable strategy: (1) Compute discriminant. (2) Positive means two distinct real. (3) Solve via factoring or formula. For example, x squared - 3x + 2 = (x-1)(x-2). You're doing amazingly—practice more!

Question 6

A quadratic with real coefficients (of the form $ax^2+bx+c=0$ ) always has exactly 2 complex solutions counting multiplicity. If its discriminant is negative, the solutions are non-real and come in complex conjugate pairs. If $2+3i$ is a solution of such a quadratic, what must the other solution be?

$-2-3i$
$2+3i$ (the same root again, not counted twice)
$-2+3i$
$2-3i$ (correct answer)

Question 7

Solve $x^2-8x+16=0$ in the complex numbers. Recall: every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity. When $b^2-4ac=0$ , there is one repeated real solution with multiplicity 2. Which answer correctly lists the solution(s) and multiplicity?

$x=4$ with multiplicity 2 (so 2 solutions counting multiplicity) (correct answer)
$x=4$ (only one solution total)
$x=4$ and $x=-4$ (two distinct real solutions)
$x=4\pm 4i$ (two complex solutions)

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex solutions counting multiplicity, especially emphasizing repeated roots. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! For x² - 8x + 16 = 0, discriminant = 64 - 64 = 0, so x = 8/2 = 4 with multiplicity 2, or (x - 4)² = 0. Choice A correctly includes multiplicity for the count of 2. Choice B is a tempting distractor treating it as only one total, ignoring multiplicity— but FTA requires counting repeats! For zero discriminant, recognize the repeated root and count it twice. Factor to see the multiplicity directly. Super work—you're nailing multiplicity!

Question 8

A quadratic with real coefficients has the form $ax^2+bx+c=0$ . In the complex numbers, it always has exactly 2 solutions counting multiplicity, and if it has a nonreal complex solution then the solutions occur in conjugate pairs. If $2+3i$ is a solution of a quadratic with real coefficients, what must the other solution be (so that there are exactly 2 complex solutions counting multiplicity)?

$2-3i$ (correct answer)
$-2+3i$
$-2-3i$
$2+3i$ (the same root must repeat)

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex solutions counting multiplicity, and for real coefficients, nonreal solutions come in conjugate pairs. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! If 2 + 3i is a solution to a real-coefficient quadratic, the other must be its conjugate 2 - 3i to ensure real coefficients and exactly 2 solutions. Choice A correctly identifies this conjugate pair principle. Choice D is a tempting distractor because it suggests the same root repeats, but that's for multiplicity in real repeated roots, not for distinct complex ones; complexes must pair as conjugates. To find the other root, simply flip the sign of the imaginary part—it's that straightforward due to coefficient symmetry. Remember, this only applies to real coefficients; if coefficients were complex, pairs aren't guaranteed. You're making excellent progress—keep applying these rules!

Question 9

Which statement best matches the "Fundamental Theorem for quadratics" idea: a degree-2 polynomial $ax^2+bx+c$ has exactly 2 complex zeros counting multiplicity, with the discriminant $b^2-4ac$ determining whether they are 2 distinct real, 1 repeated real, or 2 complex conjugates? (This also previews the Fundamental Theorem of Algebra: degree $n$ has $n$ complex zeros counting multiplicity.)

A quadratic can have 0, 1, 2, or 3 complex solutions depending on the discriminant.
A quadratic has exactly 1 solution if it touches the $x$ -axis and 2 solutions only if it crosses.
A quadratic always has exactly 2 complex solutions counting multiplicity; the discriminant only changes the type (real distinct, repeated, or complex conjugates). (correct answer)
A quadratic always has exactly 2 real solutions, because complex numbers are not needed.

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. The core idea is that over complexes, every quadratic has precisely 2 zeros counting multiplicity, with discriminant just specifying the flavor. Choice B correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies how discriminant affects type, not count. Choice A says 0,1,2, or 3 complex solutions depending on discriminant—this overstates variability; always exactly 2 total (real count as complex with zero imaginary part). To apply this transferable strategy: (1) Remember the theorem guarantees n roots for degree n. (2) Use discriminant for type. (3) Count with multiplicity. For cubics, always 3, etc. You've got this—super progress!

Question 10

Show that $x^2-4x+5=0$ has exactly 2 complex solutions counting multiplicity (as every quadratic $ax^2+bx+c=0$ does in $\mathbb{C}$ ). Use the discriminant $b^2-4ac$ to determine the type, then find the solutions.

Which option is correct?

Two complex conjugate solutions: $x=2+i$ and $x=2-i$ (correct answer)
Two distinct real solutions: $x=2\pm 1$
One repeated real solution: $x=2$ with multiplicity 2
Exactly one complex solution: $x=2+i$

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. For x² - 4x + 5 = 0, let's verify exactly 2 complex solutions: (1) Calculate discriminant: b² - 4ac = 16 - 20 = -4 (negative). (2) This means 2 complex conjugate solutions (not real). (3) Using quadratic formula: x = (4 ± √-4)/2 = (4 ± 2i)/2 = 2 ± i. (4) Solutions: 2 + i and 2 - i (two complex conjugates). Count: 2 solutions. Choice A correctly recognizes that the negative discriminant yields two complex conjugate solutions 2 + i and 2 - i. Choice B incorrectly claims positive discriminant and real solutions 2 ± 1 = 3 and 1, but we can verify these don't work: 3² - 4(3) + 5 = 9 - 12 + 5 = 2 ≠ 0. Choice C incorrectly claims zero discriminant and repeated real solution. Choice D violates the Fundamental Theorem by claiming only one complex solution—complex solutions of real-coefficient quadratics must come in conjugate pairs! Verification technique: (1) Always check your solutions by substitution. (2) For x = 2 + i: (2 + i)² - 4(2 + i) + 5 = 4 + 4i - 1 - 8 - 4i + 5 = 0 ✓. (3) Complex arithmetic: (a + bi)² = a² + 2abi + (bi)² = a² - b² + 2abi. (4) Both solutions must check out for the answer to be correct. Complex conjugate theorem: For polynomials with real coefficients, complex roots come in conjugate pairs. This maintains real coefficients when the polynomial is expanded from its factored form!

Question 11

Use the discriminant to determine the solution type for $x^2-4x+5=0$ , then find all solutions in the complex numbers. Remember: every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity (2 distinct real, 1 repeated real counted twice, or 2 complex conjugates).

One real solution: $x=2$ with multiplicity 2
Two complex solutions: $x=2+i$ and $x=2-i$ (correct answer)
No complex solutions because the discriminant is negative
Two real solutions: $x=2\pm 1$

Question 12

A quadratic with real coefficients (of the form $ax^2+bx+c=0$ ) always has exactly 2 complex solutions counting multiplicity. Also, if one non-real complex solution is $a+bi$ , the other must be its conjugate $a-bi$ .

If $x^2-4x+13=0$ has a solution $2+3i$ , what is the other solution?

$2-3i$ (correct answer)
$-2-3i$
$-2+3i$
$2+3i$ (the same value again, multiplicity 2)

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always! For quadratics with real coefficients, complex solutions must come in conjugate pairs: if a + bi is a solution, then a - bi must also be a solution. This conjugate symmetry is a direct consequence of having real coefficients—the imaginary parts cancel when substituted back into the equation. Given that x² - 4x + 13 = 0 has solution 2 + 3i, the other solution must be its conjugate 2 - 3i. We can verify: if we expand (x - (2 + 3i))(x - (2 - 3i)), we get x² - 4x + 13, confirming these are the two solutions. Choice A correctly identifies 2 - 3i as the conjugate of 2 + 3i, giving us our required pair of complex solutions. Choice B gives -2 - 3i, which changes both the real part's sign and the imaginary part's sign—that's not a conjugate! Choice C gives -2 + 3i, which only changes the real part—conjugates only flip the imaginary part's sign. Choice D incorrectly suggests the same solution appears twice, but complex solutions of real-coefficient quadratics cannot have multiplicity unless they're real (which 2 + 3i isn't). Conjugate pair recognition: (1) For real-coefficient quadratics, non-real complex solutions come in conjugate pairs. (2) If one solution is a + bi, the other is a - bi (same real part, opposite imaginary part). (3) This ensures that when multiplied out, the imaginary terms cancel, leaving real coefficients. (4) You can verify by expanding (x - (a + bi))(x - (a - bi)) = x² - 2ax + (a² + b²). Quick conjugate check: to find the conjugate of any complex number, simply change the sign of the imaginary part only. So 2 + 3i becomes 2 - 3i, and -5 - 7i becomes -5 + 7i!

Question 13

For $ax^2+bx+c=0$ , the discriminant $b^2-4ac$ determines the solution type (2 distinct real, 1 repeated real with multiplicity 2, or 2 complex conjugates), but every quadratic has exactly 2 complex solutions counting multiplicity. For $x^2+2x+5=0$ , which choice correctly describes the solutions?

One repeated real solution $x=-1$ with multiplicity $2$ .
Two complex conjugate solutions $x=-1+2i$ and $x=-1-2i$ . (correct answer)
Two distinct real solutions, since $b^2-4ac=16$ .
No solutions, because the discriminant is negative.

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. For x squared + 2x + 5 = 0, discriminant = 4 - 20 = -16 (negative), so two complex conjugates: x = (-2 ± sqrt(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. Choice C correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the conjugate pair from the negative discriminant. Choice D says no solutions because discriminant is negative—this is true over reals but not complexes; complexes ensure exactly 2 solutions always, here complex ones. To apply this transferable strategy: (1) Calculate discriminant. (2) If negative, solutions are p ± qi with p = -b/(2a), q = sqrt(|disc|)/(2a). (3) Count 2 total. For example, x squared + 1 = 0 has ±i. Excellent work—stay confident!

Question 14

A key fact (sometimes called the Fundamental Theorem for quadratics) is that any quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity: two distinct real, one repeated real (multiplicity 2), or two complex conjugates. Which choice lists one example for each case and matches it with the correct solution type determined by the discriminant $b^2-4ac$ ?

No set of three quadratics can cover all cases because quadratics can have 0 or 3 complex solutions
$x^2-5x+6=0$ (complex conjugates); $x^2-4x+4=0$ (two distinct real); $x^2+4=0$ (repeated real)
$x^2-5x+6=0$ (two distinct real); $x^2-4x+4=0$ (repeated real); $x^2+4=0$ (complex conjugates) (correct answer)
$x^2-5x+6=0$ (repeated real); $x^2-4x+4=0$ (complex conjugates); $x^2+4=0$ (two distinct real)

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex solutions counting multiplicity, with three cases based on the discriminant. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions (like x = 2 and x = 3 for x² - 5x + 6), (2) if zero, one real solution with multiplicity 2 (like x = 2 for x² - 4x + 4), (3) if negative, two complex conjugate solutions (like ±2i for x² + 4). But in ALL three cases, counting multiplicity gives exactly 2 total! Choice A correctly matches each example to its type. Choice D is a tempting distractor suggesting quadratics can have 0 or 3 solutions, but FTA fixes it at exactly 2 counting multiplicity—no more, no less! Classify by computing discriminants: positive for distinct reals, zero for repeated, negative for conjugates. Use this to pick examples for each case. You're doing wonderfully—keep exploring these!

Question 15

Factor $x^2+25$ over the complex numbers and state how many zeros it has counting multiplicity. (A degree-2 polynomial has exactly 2 complex zeros counting multiplicity; this is a special case preview of the Fundamental Theorem of Algebra: every polynomial of degree $n$ has exactly $n$ complex zeros counting multiplicity.)

$x^2+25=(x+5)(x-5)$ , so zeros are $5$ and $-5$ (2 total)
$x^2+25=(x-5i)(x+5i)$ , so zeros are $5i$ and $-5i$ (2 total) (correct answer)
$x^2+25$ cannot be factored, so it has 0 zeros
$x^2+25=(x+5i)^2$ , so there is one zero $-5i$ (multiplicity 1)

Question 16

Over the real numbers, a quadratic can have 0, 1, or 2 real solutions depending on the discriminant. Over the complex numbers, a quadratic $ax^2+bx+c=0$ always has exactly 2 solutions counting multiplicity (either 2 distinct real, 1 repeated real counted twice, or 2 complex conjugates). Which statement is correct?

A quadratic always has exactly 2 complex solutions counting multiplicity, regardless of the discriminant (correct answer)
A quadratic can have 3 complex solutions if it factors
A quadratic always has exactly 1 complex solution because complex numbers combine conjugates into one
A quadratic can have 0 complex solutions if $b^2-4ac<0$

Question 17

Solve $x^2+2x+5=0$ in the complex numbers. Recall: every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity, and the discriminant $b^2-4ac$ determines whether they are 2 distinct real, 1 repeated real (multiplicity 2), or 2 complex conjugates.

Two complex solutions: $x=-1+2i$ and $x=-1-2i$ (correct answer)
Two real solutions: $x=-1\pm 2$
One real solution: $x=-1$ (multiplicity 2)
No solutions because $b^2-4ac<0$

Question 18

Factor the quadratic $x^2+25$ over the complex numbers. Recall: every quadratic $ax^2+bx+c=0$ has exactly 2 complex zeros counting multiplicity (two real, one repeated real, or a complex conjugate pair), and this idea previews the Fundamental Theorem of Algebra: a degree- $n$ polynomial has exactly $n$ complex zeros counting multiplicity. Which factorization and zero count is correct for $x^2+25$ ?

$x^2+25=(x+5)(x-5)$ ; zeros $\pm 5$ (2 real zeros)
$x^2+25=(x+5i)(x-5i)$ ; zeros $\pm 5i$ (2 complex zeros) (correct answer)
$x^2+25$ is irreducible over the complex numbers, so it has 0 zeros
$x^2+25=(x+25)(x-1)$ ; zeros $-25$ and $1$

Explanation: This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex zeros counting multiplicity, allowing factorization over complexes even if not over reals. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! For x² + 25, discriminant = 0 - 100 = -100 < 0, so zeros ±5i, factoring as (x + 5i)(x - 5i). Choice B correctly gives this factorization and zero count. Choice C is a tempting distractor claiming irreducibility over complexes with 0 zeros, but FTA guarantees factorization into linears over complexes—always 2 for quadratics! To factor, solve x² + 25 = 0 → x = ±5i, then write (x - r1)(x - r2). This works for any quadratic—try it with positives or zeros too. Keep up the great work—you're building strong skills!

Question 19

The quadratic $f(x) = ax^2 + bx + c$ has roots $r_1 = 2 + 3i$ and $r_2 = 2 - 3i$ . If $f(0) = 39$ , what is the value of the leading coefficient $a$ ?

$a = 3$ (correct answer)
$a = 13$
$a = -3$
$a = -13$

Explanation: Since the roots are 2+3i and 2-3i, we can write f(x) = a(x-(2+3i))(x-(2-3i)) = a((x-2)-3i)((x-2)+3i) = a((x-2)² + 9) = a(x² - 4x + 4 + 9) = a(x² - 4x + 13). Since f(0) = 39, we have a(13) = 39, so a = 3. Choice B incorrectly uses a = 13 from the constant term. Choice C uses a = -3 from sign error. Choice D combines both errors.

Question 20

A quadratic polynomial $P(x) = x^2 + bx + c$ has the property that both roots lie on the circle $|z| = 3$ in the complex plane. If one root is $z_1 = -1 + 2\sqrt{2}i$ , what is the value of $c$ ?

$c = 6$
$c = 8$
$c = 9$ (correct answer)
$c = 12$

Explanation: First verify that z₁ lies on |z| = 3: |−1 + 2√2i|² = 1 + 8 = 9, so |z₁| = 3 ✓. Since P(x) has real coefficients, the other root is z₂ = −1 − 2√2i. We can verify |z₂| = 3 as well. Using Vieta's formulas, c = z₁z₂ = (−1 + 2√2i)(−1 − 2√2i) = 1 + 8 = 9. Choice A gives c = 6, which would mean |z₁z₂| = 6 ≠ 9. Choice B gives c = 8. Choice D gives c = 12.