1. **State the problem:** We want to prove that the expression $$8 - x^2 + 3x - 9 + 5(x - 1) - x(6 - 5) = 0$$ for all values of $x$. 2. **Write down the expression clearly:** $$8 - x^2 + 3x - 9 + 5(x - 1) - x(6 - 5)$$ 3. **Apply the distributive property:** $$5(x - 1) = 5x - 5$$ $$x(6 - 5) = x \times 1 = x$$ 4. **Substitute these back into the expression:** $$8 - x^2 + 3x - 9 + (5x - 5) - x$$ 5. **Combine like terms carefully:** - Constants: $8 - 9 - 5 = 8 - 14 = -6$ - $x$ terms: $3x + 5x - x = (3 + 5 - 1)x = 7x$ - $x^2$ term: $-x^2$ So the expression becomes: $$-x^2 + 7x - 6$$ 6. **Check if this expression equals zero:** We want to see if $$-x^2 + 7x - 6 = 0$$ for all $x$. This is a quadratic equation. 7. **Factor the quadratic:** $$-x^2 + 7x - 6 = 0 \implies x^2 - 7x + 6 = 0$$ (multiply both sides by $-1$) 8. **Factor:** $$x^2 - 7x + 6 = (x - 6)(x - 1) = 0$$ 9. **Solve for $x$:** $$x - 6 = 0 \implies x = 6$$ $$x - 1 = 0 \implies x = 1$$ 10. **Interpretation:** The original expression equals zero only when $x = 1$ or $x = 6$, not for all $x$. **Final answer:** The expression $$8 - x^2 + 3x - 9 + 5(x - 1) - x(6 - 5)$$ equals zero only at $x = 1$ and $x = 6$, not identically zero for all $x$.