Simplify (x² − 9)⁄(x + 3)

Correct — C. The stem is an image asking you to simplify (x² − 9)⁄(x + 3), and the numerator is a difference of two squares.

Factorise: x² − 9 = x² − 3² = (x + 3)(x − 3)
Rewrite: (x + 3)(x − 3) ⁄ (x + 3)

Cancel the common factor (x + 3), legal for x ≠ −3: x − 3
Option (c) is the image reading x − 3 → option (c).
Check with x = 5: (25 − 9)⁄8 = 2, and 5 − 3 = 2.

Why the others are wrong

• (a)
  Option (a) is the image x − 9, which keeps the 9 from x² − 9 rather than the 3 from its factor (x − 3). Cancelling leaves the root of the square, never the square itself.
• (b)
  Option (b) is the image x + 3 — the denominator copied back. That is the factor that cancels away, not the one that survives.
• (d)
  Option (d) is the image √(x² − 9), which would follow only if the whole fraction sat under a root. No radical appears anywhere in the expression.

One identity carries the question: a² − b² = (a + b)(a − b). Spotting it turns a fraction into a cancellation.

Here a = x and b = 3, so x² − 9 factors as (x + 3)(x − 3) and the denominator is one of those two factors.

Cancelling a common factor is valid wherever that factor is non-zero, which is why the simplified form x − 3 carries the restriction x ≠ −3.

Key facts

• a² − b² = (a + b)(a − b), so x² − 9 = (x + 3)(x − 3).
• (x² − 9)⁄(x + 3) = x − 3 for every x except x = −3, where the original fraction is undefined.
• Substituting one convenient value checks a simplification fast: x = 5 gives 16⁄8 = 2 = 5 − 3.

Common traps

• Cancelling the 9 against the 3 term by term, which is never legal inside a sum
• Forgetting that x = −3 must be excluded, since the original fraction is undefined there
• Starting long division when the identity finishes the job in one line

No directly related past PYQ was found.