Chapter 3 · Complex Numbers and Equations · Concept 2
3-13 Remainder Theorem (2)
Divide by something quadratic and the remainder stops being a number. It becomes ax + b — two unknowns — and you are given exactly two facts to pin them down.
Guiding question
What is left over when the divisor is bigger than a bracket?
Learning outcome. Use the Remainder Theorem to divide polynomials quickly and to solve problems.
A remainder is smaller than what you divided by
If P(x) is divided by (x − α)(x − β), call the quotient Q(x) and the remainder ax + b. Then
P(x) = (x − α)(x − β) Q(x) + ax + b
And the move that unlocks it:
(x − α)(x − β) Q(x) becomes 0 when x = α, and again when x = β. So substituting either one throws away the whole unknown part and leaves only ax + b.
Why the remainder has to be ax + b.
The degree of the remainder must be lower than the degree of the divisor. If it were not, you could divide again. The divisor here is degree 2, so the remainder is degree 1 at most — which is ax + b. Divide by a cubic and the remainder would be ax² + bx + c, needing three facts instead of two.
Q(x) is unknown. How can you throw away something you do not know? Because you are not throwing away Q — you are multiplying it by zero, and zero times anything is zero no matter how mysterious the anything is.
what we know
P ÷ (x − 1) leaves 3, P ÷ (x + 3) leaves −5. Wanted: the remainder on x² + 2x − 3.
write P in the only way that helps
P(x) = (x − 1)(x + 3) · Q(x) + ax + b
because x² + 2x − 3 = (x − 1)(x + 3), and a remainder must have lower degree than the divisor — so it is ax + b, not a number.
P(x) divided by x − 1 leaves 3, and divided by x + 3 leaves −5. Find the remainder when P(x) is divided by x² + 2x − 3.
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Factorise the divisor. Nothing works until you have done this.
x2 + 2x − 3 = (x − 1)(x + 3)And look — those are exactly the two divisors the question already told you about. That is not a coincidence; it is the question.
Work with a partner. When you have your ax + b, both of you should test it at both numbers before agreeing. It takes ten seconds and it catches every sign error.
P(x) is divisible by x − 1, and leaves a remainder of −4 when divided by x + 3. Find the remainder when P(x) is divided by x² + 2x − 3.
Writing the remainder as a single number.
Divide by a quadratic and the remainder is ax + b. A number would be the answer to a different question.
Not factorising the divisor first.
x² + x − 6 is (x − 2)(x + 3), and those brackets are what the question already told you about. Until you factorise, the two facts look unrelated to it.
Losing the sign on a.
At x = −3 the remainder is −3a + b, not 3a + b. The number you substituted brings its sign with it.
Treating "divisible" as no information.
It is the strongest fact in the question: the remainder there is exactly 0.
Not checking.
Put both numbers into your ax + b. They must give back the two remainders you started from — and this catches almost every slip.
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