Some questions give you a polynomial with unknown coefficients, such as f(x) = x³ + ax² + bx − 6, and a few facts about factors or remainders. Each fact becomes an equation in a and b, and solving those equations rebuilds the polynomial.
It combines the factor theorem and the remainder theorem with simple simultaneous equations.
What does each condition say?
There are only two kinds of condition to translate.
- “(x − a) is a factor” means f(a) = 0.
- “The remainder is R when divided by (x − a)” means f(a) = R.
A factor is really a remainder of zero. Both kinds use the same substitution. The only difference is the number on the right-hand side.
Steps for a two-unknown question
- Find the substitution value for each condition by setting the divisor to zero.
- Substitute into the polynomial and collect terms, so each condition gives one linear equation.
- Solve the two equations simultaneously.
- Write the final polynomial with the values in place.
- Check both conditions with the final polynomial.
Worked example
The polynomial f(x) = x³ + ax² + bx − 6 has (x − 1) as a factor. When f(x) is divided by (x − 3), the remainder is 12. Find a and b.
Step 1, factor condition: f(1) = 0, so 1 + a + b − 6 = 0, giving a + b = 5.
Step 2, remainder condition: f(3) = 12, so 27 + 9a + 3b − 6 = 12. That gives 9a + 3b = −9, and dividing by 3 gives 3a + b = −3.
Step 3, solve: subtract the first equation from the second: 2a = −8, so a = −4. Then b = 5 − (−4) = 9.
Step 4, polynomial: f(x) = x³ − 4x² + 9x − 6.
Step 5, checks: f(1) = 1 − 4 + 9 − 6 = 0. ✓ f(3) = 27 − 36 + 27 − 6 = 12. ✓
a = −4, b = 9
The mistake to watch for
The most common error is treating a remainder as if it were a factor.
Mistaken working: the student wrote f(3) = 0 for “remainder 12 on division by (x − 3)”, got 9a + 3b = −21, and then solved a wrong pair of equations.
The remainder is 12, not 0, so the equation must equal 12.
The correction is to write the condition in words first: “remainder 12 means f(3) = 12”. After solving, the check on the final polynomial would have shown f(3) ≠ 12 and pointed to the wrong line.
Check yourself
1. The polynomial x³ + px² + qx + 6 has factors (x − 1) and (x + 2). Find p and q.
Show answer
f(1) = 1 + p + q + 6 = 0, so p + q = −7. f(−2) = −8 + 4p − 2q + 6 = 0, so 4p − 2q = 2, which is 2p − q = 1. Add the two equations: 3p = −6, so p = −2 and q = −5. Check: f(x) = x³ − 2x² − 5x + 6, and f(1) = 0, f(−2) = −8 − 8 + 10 + 6 = 0.
2. The polynomial 2x³ + ax² + bx − 5 has (x − 1) as a factor and leaves remainder 15 when divided by (x − 2). Find a and b.
Show answer
f(1) = 2 + a + b − 5 = 0, so a + b = 3. f(2) = 16 + 4a + 2b − 5 = 15, so 4a + 2b = 4, which is 2a + b = 2. Subtract: a = −1, so a = −1, b = 4. Check: f(1) = 2 − 1 + 4 − 5 = 0 and f(2) = 16 − 4 + 8 − 5 = 15.
Where this leads next
After finding the coefficients, you can factorise with division by a linear factor and confirm the result by checking a factorisation by expansion. The mixed practice set includes questions like these, and the non-calculator working trainer is useful for the arithmetic.
If translating words into equations is where marks slip away, our teachers can practise that step with you in online one-to-one Additional Mathematics tuition.