Questions: Factoring Difference of Squares

Question 1

A student claims that 4x² − 9 cannot be factored because 4 and 9 aren't 'simple' perfect squares. What is the correct factorization?

• A. (2x − 3)(2x − 3)
• B. (4x + 3)(4x − 3)
• C. (2x + 3)(2x − 3)
• D. 4x² − 9 cannot be factored

Answer: (C)

4x² = (2x)² and 9 = 3², so a = 2x and b = 3, giving (2x + 3)(2x − 3). The key is recognizing that coefficients can be parts of perfect squares. Option A is (2x − 3)² = 4x² − 12x + 9 — a trinomial, not a difference of squares. Option B incorrectly uses 4x instead of 2x as the square root of 4x².

Question 2

Which represents the complete factorization of x⁴ − 1?

• A. (x² + 1)(x² − 1)
• B. (x + 1)²(x − 1)²
• C. (x² + 1)(x + 1)(x − 1)
• D. (x² − 1)(x² − 1)

Answer: (C)

x⁴ − 1 = (x²)² − 1² = (x² + 1)(x² − 1). The first factor is a sum of squares — it cannot be factored further over the reals. The second factor, x² − 1, is itself a difference of squares: (x + 1)(x − 1). Complete factorization means checking every factor for further factorability. Option A stops too early. Option B incorrectly treats (x² + 1) as (x + 1)².

Question 3

The expression x² + 49 can be factored as (x + 7)(x + 7).

• T. True
• F. False

Answer: False

(x + 7)(x + 7) = x² + 14x + 49, not x² + 49. A sum of squares cannot be factored over the real numbers at all — there is no pair of real binomials whose product eliminates all middle terms while producing a plus sign between the squared terms. The difference of squares pattern requires subtraction: x² − 49 = (x + 7)(x − 7).

Question 4

In the pattern a² − b² = (a + b)(a − b), the variable a should represent a single variable like x, not a compound expression like 3x or 2y³.

• T. True
• F. False

Answer: False

a and b can represent any algebraic expression. In 9x² − 25, a = 3x and b = 5, giving (3x + 5)(3x − 5). In 4y⁶ − 1, a = 2y³ and b = 1. The pattern is fully general — 'a' and 'b' are placeholders for whatever expressions, when squared, produce the two terms of the difference.

Question 5

Why does a² − b² factor into two binomials, but a² + b² cannot be factored over the real numbers?

Model answer: The factorization a² − b² = (a + b)(a − b) works because FOIL produces outer term −ab and inner term +ab, which cancel completely, leaving a² − b². For a sum a² + b², any attempt to write it as (a + c)(a − c) changes the sign of the last term, and any attempt with same-sign factors produces a nonzero middle term that can't be eliminated. The cancellation of middle terms that makes the difference factorable requires one factor to have +b and the other −b — which forces the last terms to produce a difference, not a sum.

The mechanical reason is the cancellation of FOIL's outer and inner terms. Conceptually: factoring over the reals means finding two real expressions that multiply to give the original. For the middle terms to cancel, the two factors must have opposite signs for the b-term, which forces the squared terms to subtract. There is no escape from this constraint when you need a plus sign.