Complex numbers — Question 14
Original QCE Vault practice · 5 marks
Q14 · Practice questionComplex familiar5 marks
QUESTION 14 (5 marks)
The polynomial P(z)=z4−4z3+6z2−4z−15 has real coefficients. Given that 1+2i is a root, factorise P(z) completely over C. WORKED SOLUTION
Practice marking scheme
5 marksANSWERP(z)=(z−1−2i)(z−1+2i)(z−3)(z+1). Worked solution
Since the coefficients are real, 1−2i is also a root. The conjugate pair gives (z−1−2i)(z−1+2i)=(z−1)2+4=z2−2z+5. Polynomial division gives P(z)=(z2−2z+5)(z2−2z−3). Since z2−2z−3=(z−3)(z+1), P(z)=(z−1−2i)(z−1+2i)(z−3)(z+1). Uses the conjugate-root theorem.
Forms the quadratic factor.
Determines the remaining quadratic factor.
Completes factorisation over C. Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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