What number should be subtracted from x³ − 4x² − 8x + 11 to make the number divisible by (x + 2)?

Correct — B. By the remainder theorem, the remainder on division by (x + 2) is the polynomial's value at x = −2.

f(x) = x³ − 4x² − 8x + 11
f(−2) = (−2)³ − 4(−2)² − 8(−2) + 11

= −8 − 4(4) + 16 + 11
= −8 − 16 + 16 + 11 = 3

Removing the remainder leaves an exact multiple of the divisor, so the number to subtract is 3 → option (b)

Why the others are wrong

• (a)
  9 — Subtract 9 and the value at x = −2 becomes 3 − 9 = −6, so (x + 2) still leaves a remainder. Only taking away the remainder itself clears the division.
• (c)
  5 — Subtract 5 and x = −2 gives 3 − 5 = −2. Any subtraction other than the exact remainder merely changes the remainder instead of removing it.
• (d)
  7 — Subtract 7 and x = −2 gives 3 − 7 = −4. Being close to 3 does not help — divisibility is an exact condition, not an approximate one.

The remainder theorem says the remainder of f(x) ÷ (x − a) is f(a). The divisor here is (x + 2), which is (x − (−2)), so a = −2.

f(−2) = 3 means f(x) = (x + 2)·q(x) + 3 for some quotient q(x).

Take 3 off both sides and the constant vanishes, leaving (x + 2)·q(x) — divisible by construction. That is why the number to subtract is always the remainder itself.

Sign discipline is the whole difficulty: (x + 2) sends you to −2, never to +2.

Key facts

• Remainder theorem: the remainder of f(x) ÷ (x − a) equals f(a).
• For the divisor (x + 2) the substitution is x = −2, not x = 2.
• f(−2) = −8 − 16 + 16 + 11 = 3, so 3 is both the remainder and the number to subtract.
• To force divisibility you subtract the remainder, so the factor theorem is simply the case where that remainder is already zero.

Common traps

• Substituting x = 2 for the divisor (x + 2), which returns −13 instead of 3.
• Mishandling the −8x term at x = −2, where it contributes +16 and not −16.
• Adding the remainder rather than subtracting it.

No directly related past PYQ was found.