Additional Mathematics – Factors of polynomials | e-Consult
Factors of polynomials (1 questions)
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Factor out the common factor 1 (none). Test possible rational roots ±1, ±2, ±5, ±10, ±½, ±5/2.
- Substitute x = 2: 2(8) − 5(4) − 4(2) + 10 = 16 − 20 − 8 + 10 = ‑2 → not a root.
- Substitute x = ‑1: 2(‑1)³ − 5(1) − 4(‑1) + 10 = ‑2 − 5 + 4 + 10 = 7 → not a root.
- Substitute x = 5/2: 2(125/8) − 5(25/4) − 4(5/2) + 10 = 125/4 − 125/4 − 10 + 10 = 0.
Thus x = 5/2 is a root, so (x − 5/2) is a factor. Perform polynomial division:
| 2x³ − 5x² − 4x + 10 ÷ (x − 5/2) = 2x² − 5x − 8 |
Now solve the quadratic 2x² − 5x − 8 = 0:
Discriminant Δ = (‑5)² − 4·2·(‑8) = 25 + 64 = 89.
Hence x = [5 ± √89]/4.
Final solutions: x = 5/2, x = [5 + √89]/4, x = [5 − √89]/4.