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!

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!

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. But in all cases, counting multiplicity gives exactly 2 total! This highlights the repeated root case: one distinct value but counted twice. Choice C correctly shows (x-3)²=0 with x=3 multiplicity 2, giving 2 solutions but 1 distinct. Choice D miscounts x²+1=0 as one solution i (mult1), but it's two: i and -i, distinct. Strategy: identify zero discriminant for repeats, count multiplicity as exponent in factoring. Fantastic work—you're nailing multiplicity!

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!

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!

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. But in all cases, counting multiplicity gives exactly 2 total! For a real-coefficient quadratic with one complex root like 2 + 3i, the other must be its conjugate 2 - 3i to maintain real coefficients, ensuring exactly two solutions. Choice A correctly identifies 2 - 3i as the other solution, leveraging the conjugate pair property. Choice D tempts by suggesting the same root again, but that's incorrect—conjugates are distinct unless imaginary part is zero, and multiplicity would still require proper counting. To master this, remember: for real coefficients, non-real roots come in p + qi and p - qi pairs; if you know one, flip the sign of i-term for the other. Great job exploring this—you're unlocking the beauty of complex numbers!

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!

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!

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!

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!