Extending Polynomial Identities to Complex Numbers - Algebra 2

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 $\mathbb{C}$ ?
Answer: $(x+3i)(x-3i)$ 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!
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 $b$ ?
Answer: $a = -2, b = 3$ 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 $b$ .
The identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ works for complex $a, b$ as well. Let $a = i$ and $b = 1$ . What is the value of $a^3 + b^3$ ?
Answer: $1 - i$ 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 sums like $a^3 + b^3$ 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 $a = i$ and $b = 1$ , $a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (1 + i)(-i) = 1 - i$ , matching direct computation: $(-i) + 1 = 1 - i$ . Choice A correctly applies the polynomial identity to yield $1 - i$ through precise i powers. A tempting distractor like choice B= $1 + i$ might flip the sign by miscounting $i^3$ as i instead of $-i$ ; remember the cycle: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ . To apply sum of cubes with complexes: (1) Compute $a + b$ . (2) Find $a^2$ , $-ab$ , $b^2$ and sum. (3) Multiply and simplify. Wonderful job—these examples prove identities are universal!
Use the identity $a^2-b^2=(a+b)(a-b)$ with complex numbers to compute $(1+i)^2-(2-i)^2$ by factoring. Which result is correct?
Answer: $-3+6i$ 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!
Factor completely over $\mathbb{C}$ using standard identities. $x^4+4$ (Hint: rewrite as a difference of squares: $x^4+4=(x^2+2)^2-(2x)^2$ .)
Answer: $(x^2+2x+2)(x^2-2x+2)$ 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 using clever algebraic manipulations. 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 hint suggests rewriting $x^4 + 4$ as a difference of squares. Following the hint: $x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2$ . Now apply difference of squares $a^2 - b^2 = (a+b)(a-b)$ with $a = x^2 + 2$ and $b = 2x$ : $x^4 + 4 = ((x^2+2) + 2x)((x^2+2) - 2x) = (x^2 + 2x + 2)(x^2 - 2x + 2)$ . Verify by expanding: $(x^2+2x+2)(x^2-2x+2) = x^4 - 2x^3 + 2x^2 + 2x^3 - 4x^2 + 4x + 2x^2 - 4x + 4 = x^4 + 4$ . Perfect! Choice A correctly shows the factorization as $(x^2+2x+2)(x^2-2x+2)$ . Choice B incorrectly suggests $(x^2+2)(x^2+2)$ , but this would give $x^4 + 4x^2 + 4$ , not $x^4 + 4$ . Choices C and D have wrong signs that don't produce the correct expansion. Advanced factoring technique: (1) Add and subtract a strategic middle term to create a perfect square. (2) Rewrite as difference of squares. (3) Apply the difference of squares identity. (4) Each resulting quadratic can be further factored over complex into linear factors! This Sophie Germain identity shows how creative algebraic manipulation extends our factoring power!
Over the complex numbers, you can factor using the difference of squares identity $a^2-b^2=(a+b)(a-b)$ even when $b$ involves $i$ . Factor completely over $\mathbb{C}$ : $x^2+9$
Answer: $(x+3i)(x-3i)$ 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 $i$ : $x^2 + 9 = x^2 - (-9) = x^2 - (3i)^2$ because $(3i)^2 = 9i^2 = -9$ . Now apply $a^2 - b^2 = (a+b)(a-b)$ with $a = x$ , $b = 3i$ : $x^2 + 9 = (x + 3i)(x - 3i)$ . Verify: $(x + 3i)(x - 3i) = x^2 - 3xi + 3xi - 9i^2 = x^2 - 9(-1) = x^2 + 9$ . Perfect! Choice D correctly applies the polynomial identity to complex values, factoring over complex numbers using identity patterns extended to complex domain. Choice B incorrectly factors as $(x+3)(x-3) = x^2 - 9$ , not $x^2 + 9$ —this would be the factorization of $x^2 - 9$ , a difference of squares over reals. Choice C claims it doesn't factor, forgetting that over complex numbers every sum of squares $a^2 + b^2$ factors as $(a + bi)(a - bi)$ . 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: $(a + bi)(a - bi)$ . (4) Verify factors are conjugates (important pattern!). The beauty: complex numbers don't break algebra—they extend it perfectly, allowing us to factor what was "unfactorable" over reals!
Factor $P(x)=x^2+2x+5$ over the complex numbers. (This quadratic has no real zeros, but it does factor over $\mathbb{C}$ .) Which factorization is correct?
Answer: $(x+1-2i)(x+1+2i)$ 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!
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 $\mathbb{C}$ .
Answer: $(x+3i)(x-3i)$ 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!
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 $\mathbb{C}$ .
Answer: $(x+3i)(x-3i)$ 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!
Over $\mathbb{C}$ , the difference of squares identity can be used with $b=2i$ since $(2i)^2=-4$ . Factor $x^2+4x+8$ completely over $\mathbb{C}$ .
Answer: $(x+2+2i)(x+2-2i)$ 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 quadratics with negative discriminants 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). For $x^2$ +4x +8, complete the square to $(x+2)^2$ +4 = $(x+2)^2$ - $(2i)^2$ = (x+2+2i)(x+2-2i), using difference of squares. Choice A correctly applies the polynomial identity with complex roots, yielding conjugate factors. A tempting distractor like choice D claims no factoring due to negative discriminant, but that's only over reals—complexes always allow it! To factor quadratics over C: (1) Complete the square. (2) Rewrite constant as - $(ci)^2$ . (3) Apply difference of squares. Excellent progress—this ensures every polynomial factors completely over C!
Which option correctly applies the identity $(a+b)^2 = a^2 + 2ab + b^2$ with complex numbers $a = i$ and $b = 2 - i$ to compute $(i + (2 - i))^2$ ?
Answer: $4$ 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!
Because polynomial identities extend to complex numbers, you can factor expressions that do not factor over the reals. Factor $x^2+1$ completely over $\mathbb{C}$ .
Answer: $(x+i)(x-i)$ 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). $x^2$ +1 = $x^2$ - (-1) = $x^2$ - $i^2$ , but better as $x^2$ - $(i)^2$ = (x+i)(x-i), extending difference of squares. Choice B correctly applies the polynomial identity to factor over complex numbers using conjugates. A tempting distractor like choice C is for $x^2$ -1 over reals; don't forget i for positive constants. Factoring $x^2$ +1 over complex: (1) Rewrite as $x^2$ - $i^2$ . (2) Apply (x + i)(x - i). (3) Verify by expanding back. Simple and powerful—keep practicing to master it!
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 $\mathbb{C}$ .
Answer: $x^2+16=(x+4i)(x-4i)$ 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: $x^2 + 16$ 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!
Verify the difference of squares identity over complex numbers by evaluating both sides for $a=3+2i$ and $b=1-i$ : $a^2 - b^2 \text{ and } (a+b)(a-b).$ Which option gives the common value?
Answer: $5+14i$ 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.
Over $\mathbb{C}$ , expressions that do not factor over the reals may factor using $i^2=-1$ . Which option correctly factors $x^2+4$ over the complex numbers?
Answer: $(x+2i)(x-2i)$ 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 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). The key here: $x^2$ +4 = $x^2$ - (-4) = $x^2$ - $(2i)^2$ , using difference of squares since $(2i)^2$ =-4. Choice B correctly applies the polynomial identity to factor as (x+2i)(x-2i), expanding back to $x^2$ +4 with $i^2$ =-1. Choice C confuses it with $x^2$ +1, but remember the coefficient: for $x^2$ + k, it's (x + sqrt(k) i)(x - sqrt(k) i), so for k=4, sqrt(4)=2. Factoring sums of squares over complex: (1) Identify $a^2$ + $b^2$ ; (2) Rewrite as $a^2$ - $(bi)^2$ ; (3) Factor as (a + bi)(a - bi); (4) Check conjugates. Fantastic work—you're expanding your factoring toolkit with complex numbers!
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)?
Answer: LHS: $(a+b)^2=(3)^2=9$ . RHS: $(1+i)^2+2(1+i)(2-i)+(2-i)^2=9$ . 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 familiar 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). Let's verify $(a+b)^2 = a^2 + 2ab + b^2$ 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!
The sum/difference of cubes identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ can be used with complex numbers. Let $a=1+i$ and $b=1-i$ . What is $a^3-b^3$ ?
Answer: $4i$ This question tests your understanding that polynomial identities proven for real numbers extend to complex numbers—the algebraic structures work the same way, allowing sum/difference of cubes to work with 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). For instance, with a=1+i and b=1-i, computing $a^3$ =-2+2i and $b^3$ =-2-2i gives $a^3$ - $b^3$ =4i, matching the factored form. Choice C correctly applies the polynomial identity to find $a^3$ - $b^3$ =4i, with careful complex multiplication ensuring imaginary parts add up. Choice A might result from ignoring imaginary parts or misapplying $i^2$ =-1—always expand cubes step by step, like $(a)^3$ = a * ( $a^2$ ). To use cube identities over complex: (1) Compute $a^3$ and $b^3$ separately; (2) Subtract; (3) Alternatively, factor and multiply; (4) Verify equality with organized real/imaginary tracking. You're making excellent progress—keep verifying with examples to see the consistency across number systems!
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 $\mathbb{C}$ ?
Answer: $(x+3i)(x-3i)$ 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!
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 $a$ ?
Answer: 2 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.
The identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ works for complex $a,b$ as well. Let $a=i$ and $b=1$ . What is the value of $a^3+b^3$ ?
Answer: $1-i$ 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 sums like $a^3$ + $b^3$ 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 a=i and b=1, $a^3$ + $b^3$ = (a+b)( $a^2$ - ab + $b^2$ ) = (1+i)(-i) =1-i, matching direct (-i)+1=1-i. Choice A correctly applies the polynomial identity to yield 1-i through precise i powers. A tempting distractor like choice B=1+i might flip the sign by miscounting $i^3$ as i instead of -i; remember the cycle: $i^1$ =i, $i^2$ =-1, $i^3$ =-i, $i^4$ =1. To apply sum of cubes with complexes: (1) Compute a+b. (2) Find $a^2$ , -ab, $b^2$ and sum. (3) Multiply and simplify. Wonderful job—these examples prove identities are universal!

Start this quiz