The remainder theorem says that when f(x) is divided by (x − a), the remainder is f(a). You substitute one value and calculate, instead of carrying out the whole division.
It is used whenever a question asks for “the remainder when f(x) is divided by (x + 3)” or gives a remainder and asks you to find an unknown coefficient.
Why does substitution give the remainder?
Any division can be written as f(x) = (x − a) × q(x) + R, where q(x) is the quotient and R is the remainder. Put x = a and the first term becomes zero, because (a − a) = 0. That leaves f(a) = R.
So the remainder is whatever f(a) calculates to. If f(a) = 0, there is no remainder and (x − a) is a factor, exactly as in using a known root.
How to find the value to substitute
- Set the divisor equal to zero. For (x − 2), x = 2. For (x + 3), x = −3. For (2x − 1), x = ½.
- Substitute that value into f(x). Use brackets around negative and fractional values.
- Calculate carefully and state the remainder. Write “remainder = …” so the answer is clear.
Worked example
Find the remainder when f(x) = x³ + 4x² − 2x + 5 is divided by (a) x − 2, (b) x + 3, (c) 2x − 1.
(a) x = 2: f(2) = 8 + 16 − 4 + 5 = 25.
(b) x = −3: f(−3) = (−27) + 4(9) − 2(−3) + 5 = −27 + 36 + 6 + 5 = 20.
(c) x = ½: f(½) = ⅛ + 4(¼) − 2(½) + 5 = ⅛ + 1 − 1 + 5 = 5⅛ = 41/8.
Check for (a) by long division: x³ + 4x² − 2x + 5 divided by (x − 2) gives x² + 6x + 10 with remainder 25, because (x − 2)(x² + 6x + 10) = x³ + 4x² − 2x − 20, and 5 − (−20) = 25.
The mistake to watch for
The usual slip is substituting the number that appears in the divisor rather than the value that makes it zero.
Mistaken working for (b): f(3) = 27 + 36 − 6 + 5 = 62.
The student saw ”+ 3” and substituted x = 3. The divisor (x + 3) is zero when x = −3.
The correction is one line: write “x + 3 = 0, so x = −3” before any substitution. That line takes five seconds and prevents the error.
Check yourself
1. Find the remainder when x³ − 2x + 7 is divided by (x − 3).
Show answer
f(3) = 27 − 6 + 7 = 28.
2. Find the remainder when 2x³ + x² − 5 is divided by (x + 1).
Show answer
x = −1. f(−1) = 2(−1) + 1 − 5 = −2 + 1 − 5 = −6.
3. When x³ + kx + 4 is divided by (x − 2) the remainder is 10. Find k.
Show answer
f(2) = 8 + 2k + 4 = 12 + 2k = 10, so 2k = −2 and k = −1. Check: 8 − 2 + 4 = 10.
Where this leads next
When the remainder is not zero, the division still has a quotient.
Dividing a cubic by a linear factor builds that skill, and reconstructing a polynomial from conditions uses remainders to find unknown coefficients. Try the mixed practice set afterwards. The non-calculator working trainer is handy for checking fraction arithmetic.
If remainder questions go wrong because of small sign or fraction slips, a teacher in online one-to-one Additional Mathematics tuition can spot them quickly in your written working.