If a polynomial leaves the same remainder when divided by two linear expressions, evaluate it at the zero of each divisor and set the two results equal. For a divisor such as 2x − 3, the substitution is x = 3/2.
The 2027 SEC G3 Additional Mathematics syllabus (K341) includes the remainder and factor theorems. The exercise below shows how to use the theorem even when the common remainder is not supplied.
- Set each divisor equal to zero.
- Evaluate the polynomial at both values.
- Equate the results and check the common remainder.
Worked example | Find the coefficient and remainder
Let f(x) = 2x² + kx + 5. It leaves the same remainder when divided by 2x − 3 and by x + 1. Find k and the remainder.
The zeros of the divisors are 3/2 and −1. Evaluating gives f(3/2) = 19/2 + 3k/2 and f(−1) = 7 − k. Equating them gives 19 + 3k = 14 − 2k after multiplying both sides by two. Therefore k = −1.
The common remainder is f(−1) = 7 − (−1) = 8. Check f(3/2) = 19/2 − 3/2 = 8. The zero of each divisor is a substitution value. The remainder is what the polynomial returns at that value.
Why the coefficient of x in the divisor matters
When you divide by 2x − 3, write f(x) = (2x − 3)q(x) + r. Substituting 3/2 makes the product containing q(x) vanish. Substituting 3 leaves the divisor equal to 3 and does not isolate r.
The multiplier on x affects where the divisor is zero. It does not require you to multiply the resulting remainder.
Try equal remainders and then opposite remainders
For g(x) = x² + ax + 4, compare division by 3x − 6 and x + 1. The substitutions are 2 and −1. Equal remainders give 8 + 2a = 5 − a, so a = −1 and both remainders are 6.
If the question says the first remainder is the negative of the second, set 8 + 2a = −(5 − a). This gives a = −13. The two remainders are then −18 and 18. Underline “equal” or “opposite” before forming the equation.
Frequently asked questions
Does the common remainder have to be zero?
It can be any constant. If it is zero, both divisors are factors.
What if the divisor is quadratic?
The remainder may be linear. The constant-remainder approach here is for linear divisors.
Official information
SEAB’s 2027 SEC G3 Additional Mathematics syllabus (K341), Topic A4 Polynomials and Partial Fractions. Checked 9 November 2026.