Chapter 3 · Complex Numbers and Equations · Concept 2
3-12 Remainder Theorem (1)
Economists fit polynomials to unemployment and income data, then ask what the model says about year 5. They do not divide anything — they put 5 in. This lesson is why that works.
Guiding question
How can polynomial theorems let you test a value without doing the division?
Learning outcome. Use the Remainder Theorem to divide polynomials quickly and to solve problems.
The remainder is just a value of P
The remainder when P(x) is divided by x − a is
P(a)
And the remainder when P(x) is divided by ax − b is
P(ba)
Where it comes from — two lines.
P(x) = (x − a) × Q(x) + R
P(a) = (a − a) × Q(a) + R = 0 × Q(a) + R = R
Putting x = a makes the bracket zero, so the whole quotient part vanishes no matter how complicated Q is — and only the remainder is left. That single idea runs through the next two lessons as well.
Examples, straight from the book's own page.
divided by x − 4 → P(4) · divided by 4x + 3 → P(−34)
Almost nobody gets the theorem wrong. What people get wrong is the number they put in — and the book names it: for 3x − 2, students substitute 2. Try each divisor below.
dividing by
3x − 2
Which number do you put into P to get the remainder?
1. Find the remainder when 3x⁴ − 8x³ − 5x² + 12 is divided by 3x − 2.
- 1
Find the number that makes the divisor zero.
3x − 2 = 0 → x = 23Not 2. Add the 2 across, then divide by the 3 — both steps.
Work with a partner. Before either of you substitutes anything, both write down the number that makes the divisor zero. If you have written different numbers, sort that out first — everything after it depends on it.
1–4. Find the remainder.
Question 3 comes out as zero.
That is not a mistake — a remainder of zero means 3x − 2 is a factor of 6x³ − x² + x − 2. That idea becomes the whole of lesson 3-16.
5–7. Find the value of a.
A researcher models the income distribution of a community with
P(x) = 2x3 − 3x2 + ax + 6
where x is the number of years since a new policy started. For the model to be stable it must be divisible by (2x + 1). Find a — and then think about what changes if the divisor were (x + 2) instead, and whether one polynomial can be divisible by both at once.
Dividing 3x⁴ − 8x³ − 5x² + 12 by 3x − 2 the long way takes about fifteen lines. Substituting takes three. So why is long division still taught?
Ask what each method gives you at the end.
Substituting 2 for the divisor 3x − 2.
The book names this one specifically. Set the divisor to zero: 3x − 2 = 0 gives x = 2/3, not 2.
Not flipping the sign.
For x + 3 the number is −3. The divisor has to be ZERO, and +3 makes it 6.
Powering only the top of a fraction.
(2/3)³ is 8/27, not 8/3. The bottom gets the power too.
Losing a sign inside an even power.
a(−3)² is +9a. The square kills the minus, so the term is positive.
Forgetting a missing term.
3x⁴ − 8x³ − 5x² + 12 has no x term. Leave a gap for it rather than sliding the 12 across.
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