Obtain the zeroes of the polynomial $7x^2 + 18x - 9$. Hence, write a polynomial each of whose zeroes is twice the zeroes of given polynomial.
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
Finding zeroes of $7x^2 + 18x - 9$:
Splitting the middle term:
$$7x^2 + 18x - 9 = 7x^2 + 21x - 3x - 9 = 7x(x + 3) - 3(x + 3) = (7x - 3)(x + 3)$$
Zeroes: $7x - 3 = 0 \Rightarrow x = \dfrac{3}{7}$ and $x + 3 = 0 \Rightarrow x = -3$
So the zeroes are $\alpha = \dfrac{3}{7}$ and $\beta = -3$.
New polynomial whose zeroes are twice the above:
New zeroes: $2\alpha = \dfrac{6}{7}$ and $2\beta = -6$
Sum of new zeroes $= \dfrac{6}{7} + (-6) = \dfrac{6 - 42}{7} = -\dfrac{36}{7}$
Product of new zeroes $= \dfrac{6}{7} \times (-6) = -\dfrac{36}{7}$
Required polynomial $= x^2 - \left(-\dfrac{36}{7}\right)x + \left(-\dfrac{36}{7}\right) = x^2 + \dfrac{36}{7}x - \dfrac{36}{7}$
Or equivalently: $\mathbf{7x^2 + 36x - 36}$
Source: Chapter 2, Section 2.3
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Explanation
- Splitting the middle term is the standard method to find zeroes; examiners expect you to show the factorisation step clearly.
- For the new polynomial, the zeroes are simply $2\alpha$ and $2\beta$. Use Sum and Product to build the polynomial: $k[x^2 - (\text{sum})x + (\text{product})]$.
- Multiplying through by 7 to clear fractions gives the cleaner form $7x^2 + 36x - 36$, which is acceptable (any scalar multiple is valid).
- Allocate roughly 1 mark for finding zeroes, 1 for new sum/product, 1 for the final polynomial.