This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, allowing expansions like (a+b)^2 with complex 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). Here, with a=2, b=3i, it's $4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i$. 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!

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, 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!

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 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!

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: $x^2 + 4$ 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!

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 higher degrees like $x^4$ +4 using sum of squares. 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). Treat it as $(x^2$$)^2$ + $2^2$, rewriting as $(x^2$$)^2$ - $(2i)^2$ for difference of squares. Choice B correctly applies the polynomial identity to factor as $(x^2$ $+2i)(x^2$ -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!

This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, allowing computations like $(a+b)^2$ 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!

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 over complexes. 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). Even though $x^2$ +4 doesn't factor over reals, over complexes it's $x^2$ - $(2i)^2$ = (x+2i)(x-2i), using difference of squares seamlessly. Choice B correctly applies the polynomial identity to factor over complex numbers with the conjugate pair. A tempting distractor like choice D confuses it with $x^2$-4, which does factor over reals; remember, sum of squares needs i to become a difference. Factoring sums of squares over complex: (1) Spot $x^2$ + k as $x^2$ + $(√k)^2$. (2) Rewrite as $x^2$ - (√k $i)^2$. (3) Factor as (x + √k i)(x - √k i). You've got this—complex factoring opens up new possibilities!

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 apply cube identities 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). To find $a^3 - b^3$ 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!