Chapter 3 · Complex Numbers and Equations · Concept 2
3-14 Remainder Theorem (3)
The same method as last lesson, with the facts hidden one layer deeper. Each thing you are told about a quadratic divisor is really two facts about P — and the question only wants two of the four.
Guiding question
How many facts is one remainder actually worth?
Learning outcome. Use the Remainder Theorem to divide polynomials quickly and to solve problems.
Nothing new — one extra unpacking step
If P(x) is divided by (x − α)(x − β), let the quotient be Q(x) and the remainder ax + b:
P(x) = (x − α)(x − β) Q(x) + ax + b
and the (x − α)(x − β) Q(x) part becomes 0 at x = α and at x = β.
The one new idea:
A remainder against a quadratic is two facts, not one. “Leaves 7x + 6 on (x + 2)(x − 3)” means P(−2) = 7(−2) + 6 and P(3) = 7(3) + 6 — you read the value off the REMAINDER, not off P, because at those two numbers they agree.
So the plan is always the same three moves. Factorise every divisor mentioned. Turn each given remainder into two values of P. Then look at the divisor you were ASKED about and take only the values at its roots.
The hard part of this lesson is not algebra — it is picking the right two facts out of four. Watch which ones light up.
what you are told
leaves 7x + 6 on x² − x − 6
what you are told
leaves 4x − 3 on x² + 2x − 3
what is asked
the remainder on x² + x − 2 = (x + 2)(x − 1)
P(x) leaves 7x + 6 when divided by x² − x − 6, and 4x − 3 when divided by x² + 2x − 3. Find the remainder when P(x) is divided by x² + x − 2.
- 1
Factorise all three divisors before anything else.
(x + 2)(x − 3), (x − 1)(x + 3), (x + 2)(x − 1)Now you can see which roots the asked divisor shares with the two you were given — and it takes one from each.
Work with a partner. Write down all four values of P before you look at what is asked. Then decide together which two the question needs — that decision is where the marks are.
What is the remainder when x57 + 57 is divided by x + 1?
Using the wrong two of the four values.
Look at the roots of the divisor you were ASKED about. In the Warm Up they are −2 and 1 — one from each given, and the other two values go unused.
Reading the value off P instead of off the remainder.
P(−2) = 7(−2) + 6. You substitute into the given REMAINDER, because that is the only part you know.
Not factorising every divisor.
Until all three are in brackets, you cannot see which roots they share — and that is the whole question.
Expecting a repeated bracket to give two values.
(x + 1)² has one root. Exercise 3 needs a different route: the remainder itself must leave x − 5 on (x + 1)².
Getting the degree of the remainder wrong.
Divide by a cubic and the remainder is ax² + bx + c — three unknowns, so you need three facts, not two.
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