Chapter 3 · Complex Numbers and Equations · Concept 2
3-9 Factorisation of Quadratic Expressions
At school you were told some quadratics 'do not factorise'. That was never quite true — they just did not factorise using the numbers you had. Now you have i, and every quadratic factorises.
Guiding question
Once complex numbers are allowed, is there any quadratic that will not split?
Learning outcome. Factorise quadratic expressions into their components in the set of complex numbers.
Solve it, then write the factors
If α and β are the two solutions of ax2 + bx + c = 0, then
ax2 + bx + c = a(x − α)(x − β)
Do not lose the a.
The book warns about this twice on one page, because it is the mistake almost everyone makes. For 2x² − 3x + 2 the answer starts with a 2 outside the brackets. Without it you have factorised x² − (3/2)x + 1, which is a different expression.
So the method is only two moves. Solve the equation — usually with the formula, since if it factorised nicely you would not need this lesson. Then write the two roots into the brackets with a minus in front of each, and put a in front.
The answers in this lesson are full of i, but the expressions they came from have none. So the i must cancel somewhere. Watch exactly where.
solving x² + 2x + 5 = 0 gives
x = −1 ± 2i
the two factors
…
group the real part together
…
that is (something) − (something), times (the same two) +
…
and i² is −1, so the minus becomes a plus
…
open the square
…
(a) Factorise 2x² − 3x + 2 over the complex numbers.
- 1
Ignore the factorising for a moment and just SOLVE 2x² − 3x + 2 = 0.
a = 2, b = −3, c = 2The rule needs the two roots, so finding them is the real work.
Work with a partner. Once you have an answer, one of you multiplies the brackets back out while the other watches. If the original expression does not come back, the factorisation is wrong — and this check takes less time than redoing the question.
Factorise over the complex numbers.
Not everything here needs i.
(b), (f) and (h) have positive discriminants, so their factors are all real. The instruction “over the complex numbers” does not promise that i will appear — it only removes the excuse that an expression cannot be split.
An engineer is designing a suspension bridge and models the stress on one cable. The stress depends on the square of the displacement x, and the balance of forces is
S(x) = 6x4 − 7x2 − 3
Find the possible values of x, and say what they mean for the cable under load.
You were once told that x² + 1 'cannot be factorised'. Was that a lie?
Ask what the sentence left out.
Dropping the a.
2x² − 3x + 2 factorises to 2(x − …)(x − …). Without the 2 the answer is a different expression. The book warns about this twice.
Getting the sign in the bracket wrong.
A root of α gives the factor (x − α). If α = −1 + 2i, the factor is (x + 1 − 2i) — both signs flip.
Stopping halfway on a quartic.
x⁴ + 2x² − 3 = (x² − 1)(x² + 3) is not finished. Degree 4 means four linear factors.
Expecting i to appear.
If D is positive the factors are real. "Over the complex numbers" allows i, it does not ask for it.
Not checking.
Multiply the brackets back out. Real factorisations rebuild the original in one line, and the check catches nearly every slip.
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