Remainder Theorem: Finding an Unknown Constant

The remainder theorem says: dividing $f(x)$ by $(x - a)$ leaves remainder $f(a)$. When the polynomial hides a constant, you turn that fact into an equation, set $f(a)$ equal to the stated remainder and solve. The only trap is the sign of $a$.

The polynomial $f(x) = x^3 + ax^2 - x + 6$ leaves a remainder of $4$ when divided by $(x - 2)$. Find the value of the constant $a$. [4]

The working

Step 1, identify the substitution. Division by $(x - 2)$ means evaluate at $x = 2$, and the result is the remainder: $f(2) = 4 \quad \text{(M1)}$

Step 2, compute $f(2)$ in terms of $a$: $f(2) = (2)^3 + a(2)^2 - (2) + 6 = 8 + 4a - 2 + 6 = 4a + 12 \quad \text{(M1, A1)}$

Step 3, set equal to the remainder and solve: $4a + 12 = 4 \;\Rightarrow\; 4a = -8 \;\Rightarrow\; a = -2 \quad \text{(A1)}$

Where the marks are won and lost

• The substitution is $x = 2$ for $(x - 2)$. Getting the sign right is the first method mark, $(x - 2) \to x = +2$.
• Keep the unknown through the arithmetic: $8 + 4a - 2 + 6 = 4a + 12$. Losing the $4a$ term (or mis-adding the constants) breaks the equation.
• Set $f(2) = 4$, the given remainder, not $f(2) = 0$. That $= 0$ would be the factor theorem, a different question.

Common mistakes

• Substituting $x = -2$ for the divisor $(x - 2)$.
• Setting $f(2) = 0$ (confusing remainder with factor).
• Arithmetic slips collecting $8 - 2 + 6 = 12$.

Full method: Remainder Theorem notes. Topic home: Factors of Polynomials pillar.