Remainder Theorem

When a polynomial $p(x)$ is divided by $(x - a)$, the remainder is simply $p(a)$, no long division required. The theorem turns a division question into a substitution question, and the substitution statement itself is the mark.

The basic move

Find the remainder when $p(x) = x^3 - 4x^2 + 2x + 7$ is divided by $(x - 3)$. Remainder $= p(3)$ (write this line) $= 27 - 36 + 6 + 7 =$ 4

Two marks: the statement “remainder $= p(3)$” (M), the value (A). Students who silently substitute still get the answer, and risk the M mark if an arithmetic slip spoils the A.

Divisors of the form $(ax - b)$

Substitute the value that makes the divisor zero: dividing by $(2x - 1) \to$ remainder $= p(\frac{1}{2})$; by $(x + 2) \to p(-2)$. The sign and the fraction are the two friction points, set the divisor to zero on paper ($2x - 1 = 0 \to x = \frac{1}{2}$) rather than juggling it mentally, especially on the non-calculator paper where $p(\frac{1}{2})$ means fraction arithmetic.

The real exam question: unknown coefficients

0606 rarely asks for a remainder straight; it gives remainders and asks for coefficients:

$p(x) = x^3 + ax^2 + bx - 6$. Divided by $(x - 1)$ the remainder is $-4$; divided by $(x + 2)$ the remainder is also $-4$. Find $a$ and $b$. $p(1) = -4$: $1 + a + b - 6 = -4 \to$ $a + b = 1$ $p(-2) = -4$: $-8 + 4a - 2b - 6 = -4 \to 4a - 2b = 10 \to$ $2a - b = 5$ Adding: $3a = 6 \to$ $a = 2, b = -1$

The method is the marks: each remainder condition $\to$ one equation (M each); solve the pair simultaneously (M); state $a$ and $b$ (A). Most of the marks exist before any final answer, and if your values come out as ugly fractions, treat that as a smoke alarm and re-check a substitution.

Common mistakes

• Substituting $+2$ for divisor $(x + 2)$
• Conditions translated into equations with the remainder on the wrong side ($p(a)$ equals the remainder)
• Arithmetic with fractions for $(ax - b)$ divisors rushed
• Using long division when one substitution does it, correct but slow, and slow costs marks elsewhere

Full topic context: Factors of Polynomials notes · the zero-remainder special case: factor theorem.