When f(m) = m⁵ + 5m⁴ − 3m + 7 is divided by (m − 2), then the remainder is ______.

Correct — D. The stem is a picture: f(m) = m⁵ + 5m⁴ − 3m + 7 divided by (m − 2).

Remainder theorem: dividing f(m) by (m − a) leaves f(a). Here (m − 2) fixes a = 2.

2⁵ = 32
5(2⁴) = 5 × 16 = 80
−3(2) = −6

32 + 80 = 112
112 − 6 = 106
106 + 7 = 113 → option (d).

Why the others are wrong

• (a)
  5 — 5 is the coefficient of m⁴ lifted straight out of the polynomial. The remainder theorem asks for the value of the whole function at m = 2, not for one of its coefficients.
• (b)
  0 — 0 would mean (m − 2) divides f(m) exactly, so that m = 2 is a root. Substituting gives 113, so 2 is not a root and the remainder is not zero.
• (c)
  7 — 7 is the constant term, which is f(0) — the remainder on dividing by m, not by (m − 2). The divisor decides which value of m you substitute.

The remainder theorem turns a polynomial long division into one substitution: divide f(m) by (m − a) and the remainder is f(a).

The reason is short. Write f(m) = (m − a)q(m) + r, where r is a constant because the divisor is linear. Setting m = a kills the first term and leaves f(a) = r. The factor theorem is the same statement when r happens to be 0.

Key facts

• Dividing a polynomial f(m) by (m − a) leaves remainder f(a).
• (m − a) is a factor of f(m) exactly when f(a) = 0, which is the factor theorem.
• For f(m) = m⁵ + 5m⁴ − 3m + 7, f(2) = 32 + 80 − 6 + 7 = 113.
• Dividing the same f(m) by (m + 2) would ask for f(−2), because m + 2 is m − (−2).

Common traps

• Substituting m = −2 because the divisor reads (m − 2). The sign flips once: (m − a) is tested at +a.
• Reading 5m⁴ as (5m)⁴ and computing 10⁴ instead of 5 × 16.
• Doing the long division in full. It reaches the same 113 but costs several minutes the substitution does not.

No directly related past PYQ was found.