1. **State the problem:** Factorise the quadratic equation $Sx^2 + 19x + 12 = 0$. 2. **Recall the factorisation formula:** For a quadratic $ax^2 + bx + c = 0$, factorisation involves finding two numbers that multiply to $a \times c$ and add to $b$. 3. **Identify coefficients:** Here, $a = S$, $b = 19$, and $c = 12$. 4. **Calculate product:** $a \times c = S \times 12 = 12S$. 5. **Find two numbers that multiply to $12S$ and add to 19:** Since $S$ is a variable, factorisation depends on $S$. We can write the middle term as $mx + nx$ where $m + n = 19$ and $m \times n = 12S$. 6. **Express factorisation:** $$Sx^2 + 19x + 12 = Sx^2 + mx + nx + 12$$ 7. **Group terms:** $$= (Sx^2 + mx) + (nx + 12)$$ 8. **Factor each group:** $$= x(Sx + m) + 1(n x + 12)$$ 9. **Since $Sx + m$ and $nx + 12$ must be equal for factorisation, set:** $$Sx + m = nx + 12$$ 10. **This implies $S = n$ and $m = 12$. Also, $m + n = 19$ means:** $$12 + S = 19 \implies S = 7$$ 11. **So, for $S = 7$, the quadratic becomes:** $$7x^2 + 19x + 12$$ 12. **Now factorise:** Find two numbers that multiply to $7 \times 12 = 84$ and add to 19: these are 12 and 7. 13. **Rewrite middle term:** $$7x^2 + 12x + 7x + 12$$ 14. **Group and factor:** $$= (7x^2 + 12x) + (7x + 12)$$ $$= x(7x + 12) + 1(7x + 12)$$ 15. **Factor out common binomial:** $$= (x + 1)(7x + 12)$$ **Final answer:** For $S = 7$, $$7x^2 + 19x + 12 = (x + 1)(7x + 12)$$ If $S$ is not 7, the quadratic cannot be factorised with integer coefficients.