1. Stating the problem: Factorise the expression $$7 - x^2 + 2(x - \sqrt{7})$$. 2. Expand the expression: $$7 - x^2 + 2x - 2\sqrt{7}$$ 3. Rearrange terms to group like terms: $$-x^2 + 2x + (7 - 2\sqrt{7})$$ 4. To factorise, rewrite the quadratic part in standard form: $$-x^2 + 2x + (7 - 2\sqrt{7}) = -(x^2 - 2x - 7 + 2\sqrt{7})$$ 5. Complete the square inside the parentheses: $$x^2 - 2x = (x - 1)^2 - 1$$ So, $$x^2 - 2x - 7 + 2\sqrt{7} = (x - 1)^2 - 1 - 7 + 2\sqrt{7} = (x - 1)^2 - 8 + 2\sqrt{7}$$ 6. Simplify the constant term: $$-8 + 2\sqrt{7}$$ does not factor nicely, so we write the expression as: $$-( (x - 1)^2 - 8 + 2\sqrt{7} )$$ 7. The expression is now factorised as much as possible: $$-(x - 1)^2 + 8 - 2\sqrt{7}$$ Final answer: $$7 - x^2 + 2(x - \sqrt{7}) = -(x - 1)^2 + 8 - 2\sqrt{7}$$