Derive the real quadratic factorisation from the four roots.

01 / Standard#Your turn

z^4-\sqrt3z^2+1

Official paper · jm02-2026 · 4(b)(i) · PDF 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Pair each root with its complex conjugate.

Hint 2

The sum of a unit conjugate pair is twice its cosine.

Worked solution

For the pair with arguments ±π/12, the sum and product give the first factor.
$(z-e^{i\pi/12})(z-e^{-i\pi/12})=z^2-2z\cos\frac\pi{12}+1$
The other pair has cosine −cos(π/12). Multiplying the two monic factors accounts for all four roots.
$z^4-\sqrt3z^2+1=\left(z^2-2z\cos\frac\pi{12}+1\right)\left(z^2+2z\cos\frac\pi{12}+1\right)$

The required product follows by pairing conjugate roots.

Checks and common pitfalls: Do not assume the half-angle value that the next part asks you to prove.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
For the pair with arguments ±π/12, the sum and product give the first factor.
$(z-e^{i\pi/12})(z-e^{-i\pi/12})=z^2-2z\cos\frac\pi{12}+1$
The other pair has cosine −cos(π/12). Multiplying the two monic factors accounts for all four roots.
$z^4-\sqrt3z^2+1=\left(z^2-2z\cos\frac\pi{12}+1\right)\left(z^2+2z\cos\frac\pi{12}+1\right)$

Think first. Reveal a hint when the class is ready.

2026 JM02