Answer: $(7x + 6)(7x - 6)$

Identify the perfect squares: $49x^2 = (7x)^2$ and $36 = 6^2$. The expression is $(7x)^2 - 6^2$, a difference of squares. Apply the formula: Using $a^2 - b^2 = (a + b)(a - b)$ with $a = 7x$ and $b = 6$: $(7x + 6)(7x - 6)$. Why distractors fail: Option A: $(7x - 6)^2 = 49x^2 - 84x + 36$, which has a middle term. Option C: Expanding gives $49x^2 - 7x + 252x - 36$, incorrect. Option D: Expanding gives $49x^2 - 294x + 6x - 36$, incorrect.

Answer: Because no real binomial product can produce a sum of two squares without a middle term

Consider possible binomial products: If we try $(a + b)(a + b) = a^2 + 2ab + b^2$ or $(a - b)(a - b) = a^2 - 2ab + b^2$, we always get a middle term $\pm 2ab$. The conjugate product $(a + b)(a - b) = a^2 - b^2$ eliminates the middle term but gives a minus sign. Why the correct answer works: No arrangement of two real linear binomials can multiply to give $a^2 + b^2$ because the middle terms either survive or the constant term becomes negative. Thus $a^2 + b^2$ is irreducible over the reals. Why distractors fail: Option A incorrectly invokes the concept of prime numbers, which applies to integers, not polynomial expressions. Option C is false; even exponents can be factored in many cases (e.g., $a^2 - b^2$). Option D is false; perfect squares appear in many factorizable forms like perfect square trinomials.

Answer: The middle terms $ab$ and $-ab$ cancel each other, leaving only $a^2 - b^2$

Expand using distribution: $(a + b)(a - b) = a \cdot a + a \cdot (-b) + b \cdot a + b \cdot (-b) = a^2 - ab + ab - b^2$. Observe the cancellation: The terms $-ab$ and $+ab$ are additive inverses and sum to zero, leaving $a^2 - b^2$. This cancellation is the key mechanism behind the difference of squares pattern. Why distractors fail: Option B describes what happens when both signs are positive, i.e., $(a + b)(a + b)$. Option C gives an incorrect reason; the cancellation is due to opposite signs, not lack of common factors. Option D describes an operation (squaring) that does not occur during standard expansion.

Answer: $(a + b)(a - b)$

Recall the difference of squares identity: The difference of squares formula states that $a^2 - b^2 = (a + b)(a - b)$. This is a fundamental algebraic identity. Why the correct answer works: Option B, $(a + b)(a - b)$, is the standard factored form. Expanding it: $(a + b)(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2$. Why distractors fail: Option A, $(a - b)^2$, expands to $a^2 - 2ab + b^2$, which includes a middle term. Option C, $(a + b)^2$, expands to $a^2 + 2ab + b^2$. Option D, $(a - b)(b - a)$, equals $-(a - b)^2 = -a^2 + 2ab - b^2$, which is not $a^2 - b^2$.

Answer: $x^2 - 9$

Identify the structure of a difference of squares: A difference of squares has the form $a^2 - b^2$: exactly two terms, both perfect squares, separated by subtraction. Why the correct answer works: $x^2 - 9$ fits the pattern because $x^2$ is a perfect square and $9 = 3^2$ is a perfect square, and they are connected by subtraction. Why distractors fail: Option A, $x^2 + 16$, is a sum of squares, not a difference. Option C, $x^2 - 5x + 6$, is a trinomial, not a binomial difference of two squares. Option D, $x^3 - 27$, is a difference of cubes, not squares.

Answer: Both methods are valid and yield the same result, but Method 1 is more efficient for this structure

Method 1: Difference of squares: $9x^2 - 64 = (3x)^2 - 8^2 = (3x + 8)(3x - 8)$. Direct and efficient. Method 2: AC method: Treating it as $9x^2 + 0x - 64$: $AC = 9 \times (-64) = -576$. We need two numbers that multiply to $-576$ and add to $0$: those are $24$ and $-24$. Splitting: $9x^2 + 24x - 24x - 64 = 3x(3x + 8) - 8(3x + 8) = (3x + 8)(3x - 8)$. Same result. Why distractors fail: Option A: The AC method can be applied to any quadratic, including binomials by using $b = 0$. Option B: The difference of squares formula works regardless of the leading coefficient. Option D: Both methods must produce the same unique factorization.

Answer: $16x^2 - 25y^2$

Check each option against the pattern: The difference of squares requires two perfect square terms separated by subtraction: $a^2 - b^2$. Why the correct answer works: $16x^2 = (4x)^2$ and $25y^2 = (5y)^2$. Both are perfect squares and they are subtracted, so $16x^2 - 25y^2 = (4x + 5y)(4x - 5y)$. Why distractors fail: Option A is a sum of squares, which does not factor over the reals. Option B: $7$ is not a perfect square integer, so while technically $x^2 - (\sqrt{7})^2$ works over the reals, it does not fit the standard integer-coefficient pattern. Option D is a perfect square trinomial: $(x - 3)^2$.

Answer: conjugate pair

Define the terminology: When $a^2 - b^2$ is factored as $(a + b)(a - b)$, the two factors differ only in the sign between the terms. This specific relationship is called a conjugate pair. Why the correct answer works: A conjugate pair refers to two binomials of the form $(a + b)$ and $(a - b)$, which are identical except for the sign of the second term. Why distractors fail: Option A, 'like pair,' is not a standard mathematical term for this relationship. Option C, 'symmetric pair,' is not the accepted term. Option D, 'reciprocal pair,' refers to numbers whose product is 1, which is unrelated.

Answer: $(9a^2 + 1)(3a + 1)(3a - 1)$

First factorization: $81a^4 - 1 = (9a^2)^2 - 1^2 = (9a^2 + 1)(9a^2 - 1)$. Factor the second factor further: $9a^2 - 1 = (3a)^2 - 1^2 = (3a + 1)(3a - 1)$. The factor $9a^2 + 1$ is a sum of squares and does not factor over the reals. Why distractors fail: Option A stops at the first step and is incomplete. Option C: $(3a + 1)^2(3a - 1)^2 = (9a^2 - 1)^2 = 81a^4 - 18a^2 + 1 \neq 81a^4 - 1$. Option D: $81a^2 \neq (9a^2)$—the exponent on $a$ is wrong, leading to an incorrect factorization.

Answer: $(x^2 + 4)(x + 2)(x - 2)$

First application of difference of squares: $x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$. Second application (nested difference of squares): $x^2 - 4 = x^2 - 2^2 = (x + 2)(x - 2)$. However, $x^2 + 4$ is a sum of squares and cannot be factored further over the reals. Complete factorization: $x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)$. Why distractors fail: Option B is only partially factored; $x^2 - 4$ can be factored further. Option C: $(x + 2)^2(x - 2)^2 = (x^2 - 4)^2 = x^4 - 8x^2 + 16$, not $x^4 - 16$. Option D: $(x^2 - 4)^2 = x^4 - 8x^2 + 16$, also incorrect.

Answer: $16x^2 - 49y^2$

Determine $a$ and $b$: If $(4x + 7y)$ is one factor, then by the difference of squares pattern, the other factor is $(4x - 7y)$, and $a = 4x$, $b = 7y$. Construct the expression: $a^2 - b^2 = (4x)^2 - (7y)^2 = 16x^2 - 49y^2$. Why distractors fail: Option A is a sum of squares, not a difference. Option C: $4x^2 - 7y^2$ uses unsquared coefficients ($4 \neq 4^2$ and $7 \neq 7^2$). Option D: $8x^2 - 14y^2$ uses doubled coefficients instead of squared ones.

Answer: The square root of 144 is 12, not 14, so the correct factorization is $(x + 12)(x - 12)$

Check the square root: $\sqrt{144} = 12$, not $14$. The classmate confused $12^2 = 144$ with $14^2 = 196$. Correct factorization: $x^2 - 144 = x^2 - 12^2 = (x + 12)(x - 12)$. Why distractors fail: Option A produces $x^2 - x + 144x - 144 = x^2 + 143x - 144$, which is incorrect. Option C is wrong because $x^2 - 144$ is a difference, not a sum. Option D is incorrect because $14^2 = 196 \neq 144$.

Answer: Because $x^4 - 1 = (x^2 + 1)(x^2 - 1)$, and $x^2 - 1$ is itself a difference of squares, while $x^4 + 1$ cannot be expressed as a difference of squares at any stage

Factor $x^4 - 1$: $x^4 - 1 = (x^2)^2 - 1^2 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1)$. The nested difference of squares allows repeated factorization. Consider $x^4 + 1$: $x^4 + 1$ cannot be written as a difference of two squares at any step. Since the difference of squares pattern never applies, it cannot be factored further using the methods covered in this topic. Why distractors fail: Option A is too vague; a negative constant does not always guarantee further factoring (e.g., $x^2 - 3$ does not factor over the integers). Option C is incorrect; $x^4 + 1$ does have real roots — the issue is that the difference of squares pattern never applies to it, not that it has no roots. Option D overgeneralizes: even exponents with negative constants do not automatically allow factoring in all cases.

Answer: $(x + 3)(x - 3)$

Set up the expression: The garden's area is $x^2$ and the flowerbed's area is $3^2 = 9$. The remaining area is $x^2 - 9$. Factor using difference of squares: $x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3)$. Why distractors fail: Option A: $(x - 3)^2 = x^2 - 6x + 9$, not $x^2 - 9$. Option C: $x(x - 9) = x^2 - 9x$, not $x^2 - 9$. Option D: $(x - 3)(x + 9) = x^2 + 6x - 27$, not $x^2 - 9$.

Answer: $25y^2 = (5y)^2$

Determine the square root: A perfect square term can be written as $(\text{something})^2$. For $25y^2$, we need $\sqrt{25} = 5$ and $\sqrt{y^2} = y$, so $\sqrt{25y^2} = 5y$. Why the correct answer works: $(5y)^2 = 5^2 \cdot y^2 = 25y^2$, confirming that $25y^2$ is a perfect square with square root $5y$. Why distractors fail: Option A: $(5y^2)^2 = 25y^4$, not $25y^2$. Option C: $(25y)^2 = 625y^2$, not $25y^2$. Option D writes $25y^2$ as a sum rather than a single squared term.

Answer: The product $(x + 7)(x - 7)$ equals $x^2 - 49$, not $x^2 + 49$; a sum of squares does not factor over the reals

Check the student's factorization: $(x + 7)(x - 7) = x^2 - 49$, not $x^2 + 49$. The student confused a sum of squares with a difference of squares. State the sum of squares exception: $a^2 + b^2$ does not factor into real linear binomials. Only $a^2 - b^2$ factors as $(a + b)(a - b)$. Why distractors fail: Option A: $(x - 7)(x + 7)$ is the same product due to commutativity and still equals $x^2 - 49$. Option C: $(x + 7)^2 = x^2 + 14x + 49$, not $x^2 + 49$. Option D: The leading coefficient does not need to exceed 1 for the pattern to apply.

Answer: No, because the expression has three terms and is actually a perfect square trinomial equal to $(2x - 3)^2$

Verify the student's claim: $(2x + 3)(2x - 3) = 4x^2 - 9$, which has no middle term. The original expression $4x^2 - 12x + 9$ has a $-12x$ term, so it is not a difference of squares. Identify the correct factorization: $4x^2 - 12x + 9 = (2x)^2 - 2(2x)(3) + 3^2 = (2x - 3)^2$. This is a perfect square trinomial. Why distractors fail: Option A ignores the middle term entirely. Option C is false; polynomials with leading coefficient greater than 1 can be factored. Option D incorrectly uses the sign of the middle term as justification.

Answer: $(10m + 9n)(10m - 9n)$

Identify perfect squares: $100m^2 = (10m)^2$ and $81n^2 = (9n)^2$. Apply the difference of squares pattern: $100m^2 - 81n^2 = (10m)^2 - (9n)^2 = (10m + 9n)(10m - 9n)$. Why distractors fail: Option B: $(10m - 9n)^2 = 100m^2 - 180mn + 81n^2$, which includes a middle term. Option C: Expanding gives $100m^2 - 50mn + 162mn - 81n^2 = 100m^2 + 112mn - 81n^2$, incorrect. Option D: $(10m + 9n)^2 = 100m^2 + 180mn + 81n^2$, also incorrect.

Answer: $2491$

Rewrite as a conjugate pair: Notice that $53 = 50 + 3$ and $47 = 50 - 3$. So $53 \times 47 = (50 + 3)(50 - 3)$. Apply the difference of squares: $(50 + 3)(50 - 3) = 50^2 - 3^2 = 2500 - 9 = 2491$. Why distractors fail: Option A ($2450$) likely results from an arithmetic error such as $50^2 - 50$. Option C ($2500$) forgets to subtract $3^2 = 9$. Option D ($2509$) adds $9$ instead of subtracting it.

Answer: $x^4 - 81$

Analyze Option A: $x^4 - 81 = (x^2)^2 - 9^2 = (x^2 + 9)(x^2 - 9)$. Then $x^2 - 9 = (x + 3)(x - 3)$. The factor $x^2 + 9$ is irreducible over the reals. This gives one irreducible quadratic and two linear factors after exactly two applications. Check the other options: Option B: $x^4 - 8x^2 + 16 = (x^2 - 4)^2 = ((x+2)(x-2))^2$. This is a perfect square trinomial, not a direct double application of difference of squares, and yields four linear factors (with multiplicity). Option C: $x^8 - 1$ requires more than two applications to fully factor. Option D: $x^2 - 81 = (x+9)(x-9)$, which needs only one application and yields no quadratic factor. Why Option A is the best fit: It satisfies all conditions: exactly two difference-of-squares steps, one sum-of-squares factor that stops the chain, and two linear factors.

Answer: $2(x + 5)(x - 5)$

Factor out the GCF first: $2x^2 - 50 = 2(x^2 - 25)$. Recognize and apply difference of squares: $x^2 - 25 = x^2 - 5^2 = (x + 5)(x - 5)$. So the complete factorization is $2(x + 5)(x - 5)$. Why distractors fail: Option B expands to $2x^2 - 2x + 50x - 50 = 2x^2 + 48x - 50$, incorrect. Option C is partially factored but not complete since $x^2 - 25$ can be factored further. Option D, while technically equivalent, uses irrational coefficients and is not the standard complete factorization over the integers.