1. **State the problem:** Factorise the expression $$a^2 + 2ab - ac - 3b^2 + 5bc - 2c^2$$. 2. **Group terms:** Group the terms to make factorisation easier: $$ (a^2 + 2ab - ac) + (-3b^2 + 5bc - 2c^2) $$ 3. **Factor each group:** - From the first group, factor out $a$: $$ a(a + 2b - c) $$ - From the second group, factorise the quadratic in $b$ and $c$: $$ -3b^2 + 5bc - 2c^2 = -(3b^2 - 5bc + 2c^2) $$ 4. **Factor the quadratic:** Find two numbers that multiply to $3 imes 2 = 6$ and add to $-5$. These are $-2$ and $-3$. Rewrite: $$ 3b^2 - 5bc + 2c^2 = 3b^2 - 2bc - 3bc + 2c^2 $$ Group: $$ (3b^2 - 2bc) - (3bc - 2c^2) $$ Factor each: $$ b(3b - 2c) - c(3b - 2c) $$ Factor out common binomial: $$ (b - c)(3b - 2c) $$ 5. **Substitute back:** $$ - (b - c)(3b - 2c) $$ 6. **Rewrite the original expression:** $$ a(a + 2b - c) - (b - c)(3b - 2c) $$ 7. **Check for common factors:** No common binomial factors between $a + 2b - c$ and $(b - c)(3b - 2c)$, so the factorisation is: $$ a(a + 2b - c) - (b - c)(3b - 2c) $$ This is the fully factorised form. **Final answer:** $$ a(a + 2b - c) - (b - c)(3b - 2c) $$