1. Stating the problem: Factorise the quadratic expression $$9x^2 + 3xy - 2y^2$$. 2. Formula and rules: To factorise a quadratic in two variables, look for two binomials of the form $$(ax + by)(cx + dy)$$ such that when expanded, they give the original expression. 3. Identify coefficients: The first term $9x^2$ suggests factors $3x$ and $3x$. 4. The last term $-2y^2$ suggests factors $2y$ and $-y$ (since the product must be negative). 5. Test the middle term: The middle term $3xy$ must be the sum of the cross products: $$3x \times (-y) + 3x \times 2y = -3xy + 6xy = 3xy$$ 6. This matches the middle term, so the factorisation is: $$(3x - 2y)(3x + y)$$ 7. Verify by expansion: $$3x \times 3x = 9x^2$$ $$3x \times y = 3xy$$ $$-2y \times 3x = -6xy$$ $$-2y \times y = -2y^2$$ Sum of middle terms: $3xy - 6xy = -3xy$, which does not match the original middle term $3xy$. So this is incorrect. Try option D: $$(3x + 2y)(3x - y)$$ Cross terms: $$3x \times (-y) = -3xy$$ $$2y \times 3x = 6xy$$ Sum: $-3xy + 6xy = 3xy$$, which matches the middle term. Expand fully: $$3x \times 3x = 9x^2$$ $$3x \times (-y) = -3xy$$ $$2y \times 3x = 6xy$$ $$2y \times (-y) = -2y^2$$ Sum middle terms: $-3xy + 6xy = 3xy$ Final factorisation: $$\boxed{(3x + 2y)(3x - y)}$$