Chapter 3 · Complex Numbers and Equations · Concept 2
3-16 Factor Theorem
A remainder of zero means the divisor went in exactly — so it is a factor. Which turns lesson 3-12 from a way of finding remainders into a way of factorising a cubic that nothing else will touch.
Guiding question
If you cannot see a factor, how do you find one?
Learning outcome. Apply the Factor Theorem to factorise polynomials.
Zero remainder means a factor
If P(x) is divisible by x − a — that is, if P(a) = 0 — then
P(x) = (x − a) Q(x)
where Q(x) is the quotient you get by dividing.
And how to find the a — the useful half.
Try ±(the divisors of the constant term), in order.
For x³ + 3x² + 2x + 6 the constant is 6, whose divisors are 1, 2, 3 and 6 — so try ±1, ±2, ±3, ±6 and nothing else.
Why only those numbers?
Suppose (x − r) is a factor and r is a whole number. Multiply the brackets back out: every term of the constant at the end has an r in it, so the constant must be a multiple of r. A number that does not divide the constant cannot possibly be a root — which is why the list is short, and why trying 5 on a polynomial ending in 6 is wasted effort.
This is what the method actually looks like — including the misses. Press through each one.
factorise
x³ + 2x² − x − 2
the constant term is -2, so the numbers worth trying are
1, -1, 2, -2
(b) Factorise x⁴ − 6x³ + 8x² + 8x − 16.
- 1
The constant is −16, so the numbers worth trying are ±1, ±2, ±4, ±8, ±16.
P(1) = 1 − 6 + 8 + 8 − 16 = −5, P(2) = 16 − 48 + 32 + 16 − 16 = 01 misses, 2 hits. Two tries — and there was never any point trying 3 or 5.
Work with a partner. Before trying anything, both of you list the numbers worth trying. Then split them — one takes the positives, one the negatives. It halves the work on every question here.
(d) and (f) stop early — on purpose.
2x² + x + 1 has D = 1 − 8 = −7, and x² − 2x + 3 has D = 4 − 12 = −8. Both are negative, so neither splits any further with real numbers. Leaving them as they are is the complete answer, not an unfinished one.
Factorise.
Four of them do not fully split.
In (j), (k), (l) and (p) the quadratic that is left has a positive discriminant that is not a perfect square — 33, 12, 12 and 13 — so it has real roots but not whole-number ones. In (q) and (r) the discriminant is negative altogether. Either way, you stop.
P(x) = x4 − 3x3 + ax2 + bx + 12
Both (x − 1) and (x + 2) are factors. Find a and b, and then all the roots of P(x) = 0.
How does breaking a polynomial into factors help you understand it — and real systems too?
Ask what a factor tells you that the expanded form does not.
Trying numbers at random.
Only ±(divisors of the constant term) can work. For a polynomial ending in 6 the whole list is ±1, ±2, ±3, ±6 — and nothing outside it is worth a single line.
Forgetting to try the same number twice.
Repeated factors are common. After dividing by (x − 2), try 2 again before moving on — it costs one line and (x − 2)² is a frequent answer.
Stopping at the cubic.
Once one factor is out, the quotient still has to be dealt with — factorised the ordinary way if it is a quadratic, or hunted again if it is a cubic.
Forcing a quadratic that will not go.
Check its discriminant. If it is negative, or positive but not a perfect square, leave the quadratic as it stands. That is the finished answer.
Losing the number in front.
For 2x³ − x² − 5x − 2 the answer is (x − 2)(2x + 1)(x + 1). The 2 has to appear somewhere, or the brackets multiply back to the wrong thing.
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