If $p$ and $q$ are zeroes of the polynomial $p(y) = 21y^2 - y - 2$, then find the value of $(1 - p) \cdot (1 - q)$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
For $p(y) = 21y^2 - y - 2$, with zeroes $p$ and $q$:
$$p + q = \frac{-(-1)}{21} = \frac{1}{21}, \qquad pq = \frac{-2}{21}$$
Now,
$$(1-p)(1-q) = 1 - (p+q) + pq = 1 - \frac{1}{21} + \frac{-2}{21}$$
$$= 1 - \frac{1}{21} - \frac{2}{21} = 1 - \frac{3}{21} = 1 - \frac{1}{7} = \frac{6}{7}$$
$$\therefore\ (1-p)(1-q) = \dfrac{6}{7}$$
Source: Chapter 2, Section 2.3
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Explanation
- Use $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$ directly — no need to find the actual zeroes.
- Expand $(1-p)(1-q) = 1-(p+q)+pq$ algebraically and substitute.
- Examiners award 1 mark for correctly finding sum and product, and 1 mark for the final computation. Show each step clearly.