Proving and Applying Polynomial Identities - Algebra 2

Prove the polynomial identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$ for all real $x, y$ by expanding both sides and simplifying. Which option correctly completes the proof?
Answer: Expand LHS: $(x^2+y^2)^2=x^4+2x^2y^2+y^4$ . Expand RHS: $(x^2-y^2)^2+(2xy)^2=(x^4-2x^2y^2+y^4)+4x^2y^2=x^4+2x^2y^2+y^4$ . Since both sides match, the identity is true for all $x,y$ . This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. For example, (a + b)² = a² + 2ab + b² works whether a = 5 and b = 3, or a = -2 and b = 7, or any other values. To prove an identity, we show the two sides are algebraically equivalent by expanding, factoring, or manipulating until they match. Let's expand both sides: LHS = (x² + y²)² = x⁴ + 2x²y² + y⁴ using the square formula. RHS = (x² - y²)² + (2xy)² = (x⁴ - 2x²y² + y⁴) + 4x²y² = x⁴ + 2x²y² + y⁴. Since both sides equal x⁴ + 2x²y² + y⁴, the identity is proven! Choice A correctly shows this expansion and simplification. Choice B makes an error by writing 2xy instead of 2x²y² in the LHS expansion—remember we're squaring (x² + y²), not (x + y). To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side. Alternatively, expand BOTH sides independently and show they give the same result. Either way, the goal is demonstrating the two expressions are algebraically equivalent for all variable values.
Apply the sum of cubes identity to factor the polynomial $x^3+8$ completely over the integers.
Answer: $(x+2)(x^2-2x+4)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. The sum of cubes identity a³ + b³ = (a + b)(a² - ab + b²) is a powerful factoring tool. To apply it to x³ + 8, we need to recognize that 8 = 2³, so we have x³ + 2³. Using the identity with a = x and b = 2: x³ + 2³ = (x + 2)(x² - x(2) + 2²) = (x + 2)(x² - 2x + 4). Choice C correctly applies the sum of cubes pattern, giving (x + 2)(x² - 2x + 4). Choice A has the wrong sign in the first factor (should be + not just +), Choice B treats 8 as if it equals 8³ instead of 2³, and Choice D uses the wrong sign pattern (that would be for difference of cubes). To apply sum/difference of cubes identities: (1) Recognize perfect cubes (8 = 2³, 27 = 3³, 64 = 4³, etc.), (2) Identify whether it's sum (+) or difference (-), (3) Apply the correct pattern: sum uses (a + b)(a² - ab + b²), difference uses (a - b)(a² + ab + b²), (4) Simplify the resulting factors. These identities are invaluable for factoring cubic expressions that would otherwise be difficult to handle!
Use the identity $(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$ to generate a Pythagorean triple when $x=4$ and $y=1$ .
Answer: $(8,15,17)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. The Pythagorean triple identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$ is remarkably powerful: it generates right triangles! If you pick any two integers x and y (with x > y), this identity gives you three expressions— $x^2 + y^2$ , $x^2 - y^2$ , and $2xy$ —that form a Pythagorean triple. For x=4, y=1: a = 16 - 1 = 15, b = 2·4·1 = 8, c = 16 + 1 = 17, and $15^2 + 8^2 = 225 + 64 = 289 = 17^2$ . Choice A correctly generates the triple (8,15,17) using the identity. Other choices have incorrect orders or values like 16 instead of 17. For generating Pythagorean triples with $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$ : (1) Pick two positive integers x and y with x greater than y, (2) Calculate the three quantities: a = $x^2 - y^2$ , b = 2xy, c = $x^2 + y^2$ , (3) Verify: $a^2 + b^2$ should equal $c^2$ .
Recognize and factor completely using polynomial identities: $x^4 - 16.$
Answer: $(x - 2)(x + 2)(x^2 + 4)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. For example, $(a + b)^2 = a^2 + 2ab + b^2$ works whether a = 5 and b = 3, or a = -2 and b = 7, or any other values. To prove an identity, we show the two sides are algebraically equivalent by expanding, factoring, or manipulating until they match. For $x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ , using difference of squares twice for complete factorization. Choice C correctly factors completely as $(x - 2)(x + 2)(x^2 + 4)$ . A distractor like Choice A stops at partial factorization, but complete means breaking down all factorable parts. To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side. Alternatively, expand BOTH sides independently and show they give the same result. Either way, the goal is demonstrating the two expressions are algebraically equivalent for all variable values. Testing with specific numbers can give confidence but doesn't prove—you need algebraic demonstration!
Prove the polynomial identity $a^2 - b^2 = (a + b)(a - b)$ by expanding the right-hand side and simplifying.
Answer: $(a + b)(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. For example, $(a + b)^2 = a^2 + 2ab + b^2$ works whether a = 5 and b = 3, or a = -2 and b = 7, or any other values. To prove an identity, we show the two sides are algebraically equivalent by expanding, factoring, or manipulating until they match. Let's expand the right side: $(a + b)(a - b) = a(a) + a(-b) + b(a) + b(-b) = a^2 - ab + ab - b^2$ , and notice $-ab + ab$ cancels out, leaving $a^2 - b^2$ , which matches the left side. Choice A correctly proves the identity through proper expansion and shows the canceling terms leading to $a^2 - b^2$ . To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side.
Demonstrate that the identity $a^2-b^2=(a+b)(a-b)$ holds for all real $a,b$ by expanding the right-hand side. Which expansion is correct?
Answer: $(a+b)(a-b)=a^2-b^2$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The difference of squares identity a² - b² = (a + b)(a - b) is one of the most useful factoring patterns in algebra. Let's expand the right side to verify: (a + b)(a - b) = a(a - b) + b(a - b) = a² - ab + ba - b² = a² - ab + ab - b² = a² - b². The cross terms -ab and +ab cancel perfectly! Choice A correctly shows that (a + b)(a - b) = a² - b². Choice B incorrectly includes a 2ab term, confusing this with the perfect square expansion. Choice C has -ab without the canceling +ab term. Choice D claims the result is a² + b², which would be the sum of squares, not the difference! To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side. The difference of squares pattern is special because the middle terms always cancel, leaving just a² - b²!
Prove the identity $(a-b)^3 = a^3 - 3a^2 b + 3 a b^2 - b^3$ by expanding $(a-b)^3$ , then apply it to expand $(x-2)^3$ .
Answer: $(x-2)^3 = x^3 - 6x^2 + 12x - 8$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The binomial cube identity $(a - b)^3 = a^3 - 3a^2 b + 3 a b^2 - b^3$ follows a specific pattern with coefficients 1, -3, 3, -1. Let's apply it to $(x - 2)^3$ with a = x and b = 2: $(x - 2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = x^3 - 6x^2 + 3x(4) - 8 = x^3 - 6x^2 + 12x - 8$ . Choice A correctly shows this expansion with all terms properly calculated. Choice B has 4x instead of 12x (forgetting to multiply 3 × 4), Choice C has - $4x^2$ instead of - $6x^2$ (using $2^2$ instead of 3 × 2), and Choice D incorrectly has -12x instead of +12x (wrong sign pattern). To expand binomial cubes: (1) Remember the coefficient pattern: 1, 3, 3, 1 for both $(a + b)^3$ and $(a - b)^3$ , (2) For $(a - b)^3$ , signs alternate: +, -, +, -, (3) Powers decrease for a (from 3 to 0) and increase for b (from 0 to 3), (4) Calculate each term carefully—it's easy to make arithmetic errors with the coefficients!
Prove the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ by expanding $(a-b)^3=(a-b)(a-b)(a-b)$ . Which expansion is correct?
Answer: $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. To prove the cube of a difference identity, we need to expand (a - b)³ = (a - b)(a - b)(a - b) systematically. First, expand (a - b)(a - b) = a² - 2ab + b². Then multiply by (a - b) again: (a² - 2ab + b²)(a - b) = a²(a - b) - 2ab(a - b) + b²(a - b) = a³ - a²b - 2a²b + 2ab² + ab² - b³ = a³ - 3a²b + 3ab² - b³. Choice A correctly shows (a - b)³ = a³ - 3a²b + 3ab² - b³. Choice B incorrectly has all negative signs in the middle terms. Choice C has wrong coefficients (2 instead of 3). Choice D shows the expansion of (a + b)³, not (a - b)³! To expand powers of binomials: (1) Use repeated multiplication or Pascal's triangle for coefficients, (2) For (a - b)ⁿ, signs alternate: +, -, +, -, ..., (3) Powers of a decrease while powers of b increase, (4) Check your work by testing with simple values like a = 2, b = 1.
Apply the difference of cubes identity to factor the polynomial completely: $x^3 - 8$ .
Answer: $(x-2)(x^2+2x+4)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. The difference of cubes identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ is remarkably powerful for factoring! For $x^3 - 8 = x^3 - 2^3$ , we identify a = x and b = 2, so it factors as $(x - 2)(x^2 + x \cdot 2 + 2^2) = (x - 2)(x^2 + 2x + 4)$ . Note the signs: it's +ab in the quadratic for difference of cubes, unlike sum of cubes which has -ab. Choice C correctly applies the difference of cubes identity with the proper signs and terms. Choices A and B have wrong signs in the quadratic, and D isn't using the identity properly. To factor using identities: (1) Recognize the pattern like sum or difference of cubes, (2) Identify a and b, (3) Plug into the formula carefully, especially the signs, (4) Verify by expanding back if needed.
Verify that the polynomial identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ is true for all real $a,b$ by expanding the right-hand side.
Answer: $(a+b)(a^2-ab+b^2)=a^3-a^2b+ab^2+a^2b-ab^2+b^3=a^3+b^3$ so the identity holds for all $a,b$ . This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The sum of cubes identity a³ + b³ = (a + b)(a² - ab + b²) is a fundamental factoring pattern that works for any values of a and b. Let's expand the right side using distribution: (a + b)(a² - ab + b²) = a(a² - ab + b²) + b(a² - ab + b²) = a³ - a²b + ab² + a²b - ab² + b³ = a³ + b³. Notice how the middle terms -a²b + a²b and +ab² - ab² cancel out perfectly! Choice B correctly shows this expansion step-by-step, demonstrating that the identity holds for all real a, b. Choice A makes an arithmetic error in combining terms, C incorrectly claims the result equals (a + b)³, and D adds an extra term ab(a + b) that suggests the identity only works sometimes. To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side. The beauty of identities like sum of cubes is that the middle terms always cancel—this isn't coincidence but careful algebraic design!
Demonstrate that the polynomial identity $(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$ holds for all real $x,y$ by expanding both sides and simplifying.
Answer: LHS: $(x^2+y^2)^2=x^4+2x^2y^2+y^4$ and RHS: $(x^2-y^2)^2+(2xy)^2=(x^4-2x^2y^2+y^4)+4x^2y^2=x^4+2x^2y^2+y^4$ so they are equal for all $x,y$ . This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. For example, (a + b)² = a² + 2ab + b² works whether a = 5 and b = 3, or a = -2 and b = 7, or any other values. To prove an identity, we show the two sides are algebraically equivalent by expanding, factoring, or manipulating until they match. Let's expand the left side: (x² + y²)² = x⁴ + 2x²y² + y⁴. Now the right side: (x² - y²)² + (2xy)² = (x⁴ - 2x²y² + y⁴) + 4x²y² = x⁴ - 2x²y² + y⁴ + 4x²y² = x⁴ + 2x²y² + y⁴. Choice B correctly shows both expansions and demonstrates they're equal, proving the identity holds for all x, y. Choice A makes an error in expanding (2xy)² as 2x²y² instead of 4x²y², while C incorrectly expands the squared binomials, and D adds an extra 4xy term that shouldn't be there. To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side. Alternatively, expand BOTH sides independently and show they give the same result. Either way, the goal is demonstrating the two expressions are algebraically equivalent for all variable values.
Recognize $x^4-16$ as a difference of squares (possibly more than once) and factor it completely over the integers using polynomial identities.
Answer: $(x-2)(x+2)(x^2+4)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. To factor x⁴ - 16, we should recognize it as a difference of squares: x⁴ - 16 = (x²)² - 4². Using a² - b² = (a + b)(a - b), we get: x⁴ - 16 = (x² + 4)(x² - 4). But wait—x² - 4 is also a difference of squares! So: x² - 4 = (x + 2)(x - 2). Therefore: x⁴ - 16 = (x² + 4)(x + 2)(x - 2) = (x - 2)(x + 2)(x² + 4). Choice B correctly factors x⁴ - 16 completely over the integers as (x - 2)(x + 2)(x² + 4). Choice A stops after one application of difference of squares, C incorrectly factors as if 16 = 16² instead of 4², and D incorrectly suggests x² + 4 can be factored over the integers (it can't—sum of squares doesn't factor over reals). To factor using difference of squares repeatedly: (1) Look for perfect squares being subtracted, (2) Apply a² - b² = (a + b)(a - b), (3) Check if any resulting factors are themselves differences of squares, (4) Continue until no more factoring is possible over your number system. This technique is powerful for factoring high-degree polynomials!
Apply the difference of cubes identity to factor $x^3-27$ completely over the integers.
Answer: $(x-3)(x^2+3x+9)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The difference of cubes identity a³ - b³ = (a - b)(a² + ab + b²) provides a direct way to factor expressions like x³ - 27. First, recognize that 27 = 3³, so we have x³ - 3³. Applying the identity with a = x and b = 3: x³ - 27 = x³ - 3³ = (x - 3)(x² + x(3) + 3²) = (x - 3)(x² + 3x + 9). Choice C correctly applies the difference of cubes pattern to factor x³ - 27 as (x - 3)(x² + 3x + 9). Choice A incorrectly has a minus sign in the middle term of the second factor (should be +3x, not -3x), choice B uses addition instead of subtraction in the first factor, and choice D doesn't follow the difference of cubes pattern at all. To apply sum or difference of cubes identities: (1) Recognize the form a³ ± b³, (2) Identify what a and b are (here x and 3), (3) Apply the pattern: sum gives (a + b)(a² - ab + b²), difference gives (a - b)(a² + ab + b²), (4) Substitute and simplify. Remember the sign patterns: difference of cubes has plus signs throughout the second factor, sum of cubes has a minus in the middle!
Prove the identity $(a - b)^3 = a^3 - 3a^2 b + 3 a b^2 - b^3$ by expanding $(a - b)^3$ , then apply it to expand $(x - 2)^3$ .
Answer: $ $(x - 2)^3$ = $x^3$ - $6x^2$ + 12x - 8 $ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The binomial cube identity $(a - b)^3 = a^3 - 3a^2 b + 3 a b^2 - b^3$ follows a specific pattern with coefficients 1, -3, 3, -1. Let's apply it to $(x - 2)^3$ with a = x and b = 2: $(x - 2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = x^3 - 6x^2 + 3x(4) - 8 = x^3 - 6x^2 + 12x - 8$ . Choice A correctly shows this expansion with all terms properly calculated. Choice B has 4x instead of 12x (forgetting to multiply 3 × 4), Choice C has - $4x^2$ instead of - $6x^2$ (using $2^2$ instead of 3 × 2), and Choice D incorrectly has -12x instead of +12x (wrong sign pattern). To expand binomial cubes: (1) Remember the coefficient pattern: 1, 3, 3, 1 for both $(a + b)^3$ and $(a - b)^3$ , (2) For $(a - b)^3$ , signs alternate: +, -, +, -, (3) Powers decrease for a (from 3 to 0) and increase for b (from 0 to 3), (4) Calculate each term carefully—it's easy to make arithmetic errors with the coefficients!
Verify that the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ is true for all real $a$ and $b$ by expanding the right-hand side.
Answer: $(a+b)(a^2-ab+b^2)=a^3-a^2b+ab^2+a^2b-ab^2+b^3=a^3+b^3$ , so the identity holds for all $a,b$ . This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The sum of cubes identity a³ + b³ = (a + b)(a² - ab + b²) is a fundamental factoring pattern that appears frequently in algebra. Let's expand the right side step by step: (a + b)(a² - ab + b²) = a(a² - ab + b²) + b(a² - ab + b²) = a³ - a²b + ab² + ba² - ab² + b³ = a³ - a²b + ab² + a²b - ab² + b³ = a³ + b³. Notice how the middle terms cancel out perfectly! Choice B correctly proves the identity through proper expansion, showing all the intermediate steps including how -a²b and +a²b cancel, as well as how +ab² and -ab² cancel. Choice A makes an error by writing -ab² instead of -a²b in the first expansion, while choices C and D have incorrect expansions that don't lead to the proper cancellation. To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side. The key insight in the sum of cubes identity is that the middle factor (a² - ab + b²) is specifically designed so that cross terms cancel when multiplied by (a + b).
Use the binomial square identity $(a+b)^2=a^2+2ab+b^2$ to compute $101^2$ mentally.
Answer: $101^2=10201$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. For example, $(a + b)^2 = a^2 + 2ab + b^2$ works whether a = 5 and b = 3, or a = -2 and b = 7, or any other values. To compute $101^2$ , rewrite as $(100 + 1)^2 = 100^2 + 2 \cdot 100 \cdot 1 + 1^2 = 10000 + 200 + 1 = 10201$ . Choice C correctly applies the binomial square identity to get 10201. Choices A, B, and D have calculation errors like missing the 200 or misadding. To use identities for mental math: (1) Break down the number, (2) Apply the pattern, (3) Compute each term, (4) Add up carefully—great job practicing this skill!
Recognize a pattern and factor completely using polynomial identities: $x^4-16.$
Answer: $(x-2)(x+2)(x^2+4)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. For example, $(a + b)^2 = a^2 + 2ab + b^2$ works whether a = 5 and b = 3, or a = -2 and b = 7, or any other values. Here, $x^4 - 16 = (x^2)^2 - 4^2$ , so using difference of squares: $(x^2 - 4)(x^2 + 4)$ , and then factor $x^2 - 4$ further as $(x - 2)(x + 2)$ , giving $(x - 2)(x + 2)(x^2 + 4)$ . Choice B correctly applies the difference of squares twice for complete factoring. Choice A stops too early without factoring $x^2 - 4$ , while C and D have incorrect constants. To factor using identities: (1) Look for patterns like difference of squares or cubes, (2) Apply step by step, (3) Check for further factoring, (4) Verify by multiplying back.
Verify that the polynomial identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ is true for all real numbers $a$ and $b$ by expanding the right-hand side.
Answer: $(a+b)(a^2-ab+b^2)=a^3-a^2b+ab^2+a^2b-ab^2+b^3=a^3+b^3$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. For example, $(a + b)^2 = a^2 + 2ab + b^2$ works whether $a = 5$ and $b = 3$ , or $a = -2$ and $b = 7$ , or any other values. To prove an identity, we show the two sides are algebraically equivalent by expanding, factoring, or manipulating until they match. Expanding the right side: $(a + b)(a^2 - ab + b^2) = a \cdot a^2 + a \cdot (-ab) + a \cdot b^2 + b \cdot a^2 + b \cdot (-ab) + b \cdot b^2 = a^3 - a^2 b + a b^2 + a^2 b - a b^2 + b^3$ , where $-a^2 b + a^2 b$ and $a b^2 - a b^2$ cancel, leaving $a^3 + b^3$ . Choice A correctly proves the identity by showing the expansion and cancellation to match $a^3 + b^3$ . To prove a polynomial identity: (1) Pick the more complicated side (usually), (2) Expand using FOIL, distribution, or combining like terms, (3) Simplify systematically, (4) Show it equals the other side.
Recognize $x^4-16$ as a difference of squares twice and factor completely using identities. Which answer is correct?
Answer: $(x-2)(x+2)(x^2+4)$ This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. To factor x⁴ - 16 completely, we need to recognize it as a difference of squares and apply the pattern repeatedly. First, x⁴ - 16 = (x²)² - 4² = (x² - 4)(x² + 4) using a² - b² = (a - b)(a + b). But we're not done! The factor x² - 4 is also a difference of squares: x² - 4 = x² - 2² = (x - 2)(x + 2). Therefore, x⁴ - 16 = (x - 2)(x + 2)(x² + 4). Note that x² + 4 cannot be factored further using real numbers. Choice B correctly shows the complete factorization (x - 2)(x + 2)(x² + 4). Choice A stops after one application of difference of squares, leaving x² - 4 unfactored. Choice C incorrectly factors as if we had x⁴ - 256 = x⁴ - 4⁴. Choice D suggests (x - 2)²(x + 2)², which would expand to x⁴ - 8x² + 16, not x⁴ - 16. For complete factoring: (1) Look for common factors first, (2) Check if it's a special pattern (difference of squares, sum/difference of cubes), (3) Apply the pattern, (4) Check if any factors can be factored further, (5) Repeat until all factors are prime or irreducible!
A student claims that the identity $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ can be used to prove that $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ . Which step in the verification process reveals whether this approach is valid?
Answer: Expanding $(x + y)(x^2 - xy + y^2)$ and checking if it equals $x^3 + y^3$ When verifying algebraic identities, you need to test whether two expressions are actually equal by expanding and comparing them directly. The student's proposed approach tries to connect the binomial expansion $(x + y)^3$ to the sum of cubes factorization, but this creates an unnecessarily complex verification path. The most straightforward way to verify the identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ is to expand the right side and see if it equals the left side. When you multiply $(x + y)(x^2 - xy + y^2)$ , you get: $x(x^2 - xy + y^2) + y(x^2 - xy + y^2) = x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3$ The middle terms cancel: $-x^2y + x^2y = 0$ and $xy^2 - xy^2 = 0$ , leaving exactly $x^3 + y^3$ . This confirms the identity is correct. Option A focuses on structural comparison rather than actual verification. Option B suggests using the binomial expansion as an intermediate step, which overcomplicates the process and doesn't directly verify the sum of cubes formula. Option C tries to match coefficients between different expansions, but the binomial theorem and sum of cubes factorization involve different algebraic structures that don't directly correspond. Option D correctly identifies that direct expansion and comparison is the clearest verification method. Remember: when verifying algebraic identities, the most reliable approach is usually to expand one side completely and check if it matches the other side exactly.

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