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The factor and remainder theorems
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Substituting one number can tell you whether a bracket is a factor, or what the remainder will be, without any long division.
The remainder theorem
When f(x) is divided by (x − a), the remainder is f(a).
When f(x) is divided by (ax − b), the remainder is f(ba).
Remainder of x3 + 4x − 1 ÷ (x − 2): f(2) = 8 + 8 − 1 = 15
When f(x) is divided by (ax − b), the remainder is f(ba).
Remainder of x3 + 4x − 1 ÷ (x − 2): f(2) = 8 + 8 − 1 = 15
The factor theorem
If f(a) = 0, then (x − a) is a factor of f(x).
f(x) = x3 − 7x + 6: f(1) = 1 − 7 + 6 = 0, so (x − 1) is a factor.
f(x) = x3 − 7x + 6: f(1) = 1 − 7 + 6 = 0, so (x − 1) is a factor.
Fully factorising a cubic
Find one factor by trying small values (±1, ±2, ±3...). Divide to get a quadratic, then factorise that.
x3 − 7x + 6 = (x − 1)(x2 + x − 6) = (x − 1)(x − 2)(x + 3)
x3 − 7x + 6 = (x − 1)(x2 + x − 6) = (x − 1)(x − 2)(x + 3)
(x − 3) is a factor of f(x) = x3 − 2x2 + ax + 6. Find a.
- f(3) = 0
- 27 − 18 + 3a + 6 = 0
- 15 + 3a = 0
Answer: a = −5
Remainder on dividing by (x − a) is f(a). If f(a) = 0, (x − a) is a factor. For (ax − b) substitute x = ba.
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