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!

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!

This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, supporting advanced factorizations. Key concept: The difference of cubes $a^3$ - $b^3$ = $(a-b)(a^2$ + 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!

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 sums of squares. Key concept: Rewrite $x^2$ +4 = $x^2$ - $(2i)^2$ using $(2i)^2$=-4, then apply difference of squares for (x+2i)(x-2i). Verify: $(x+2i)(x-2i)=x^2$ - $(2i)^2$$=x^2$ +4, spot on! Choice B correctly identifies this factorization. A tempting distractor like C fails by using 1 instead of $4=2^2$, leading to $x^2$+1 instead of +4—remember to match the constant term! Transferable strategy: Factor $x^2$ + k = (x + sqrt(k) i)(x - sqrt(k) i), verifying expansion. You're mastering complex factoring—keep going!

This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, verifying by evaluation. Key concept: Difference of squares holds, so compute $a^2 - b^2$ and $(a+b)(a-b)$ for $a=3+2i$, $b=1-i$, both yielding $5+14i$. Direct: $a^2=5+12i$, $b^2=-2i$, difference $5+14i$; factored matches. Choice A correctly identifies $5+14i$ as the common value. A tempting distractor like B flips imaginary sign, perhaps from error in a-b computation. Transferable strategy: Always compute both forms, simplify step-by-step, and equate—reinforces the identity! Keep up the superb work.

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!

This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, allowing creative factoring like adding imaginaries. Key concept: Rewrite $x^4$ +4 = $(x^2$$)^2$ +4 = $(x^2$$)^2$ - $(2i)^2$ = $(x^2$ $-2i)(x^2$ +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!

This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, enabling factoring quadratics with negative discriminants. Key concept: Quadratics factor over C using roots from quadratic formula, incorporating i for sqrt(negative). For $x^2$ +2x +5, discriminant $-16=(4i)^2$, roots -1 ± 2i, so (x +1 -2i)(x +1 +2i). Choice A correctly gives this conjugate pair. A tempting distractor like B uses ±i instead of ±2i, factoring $x^2$ +2x +2 instead—check discriminant! Transferable strategy: Solve $ax^2$ +bx +c=0 with roots [-b ± $sqrt(b^2$-4ac)]/(2a), factor as (x - root1)(x - root2). You're excelling at this!

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!

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 verify them with complex substitutions. All polynomial identities that work for real numbers also work for complex numbers because complex numbers form an algebraically closed field—addition, subtraction, multiplication follow the same properties (commutative, associative, distributive). For example, with $a=1+i$ 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!