Nomenclature: the terminology for a specific topic Quadratic: a polynomial whose highest power is 2 Coefficient: the number before the variable The Technique: Let us take a concrete example: \(x^2-3x-18\). To factorise this monic quadratic we need to find two numbers that multiply to give -18 and add to give -3. Such numbers will be -6 and 3. Thus we can write \(x^{2}-3 x-18=(x+3)(x-6)\) Examples: Factorise: \(x^{2}+8 x+16\) We are looking for two numbers that multiply to give 16 and add to give 8. The numbers 4 and 4 satisfy both conditions. Thus \(x^{2}+8 x+16=(x+4)(x+4)\) \[ =(x+4)^{2} \] \(x^{2}+5 x+6\) We are looking for two numbers that multiply to give 6 and add to give 5. The numbers 2 and 3 satisfy both conditions. Thus \(x^{2}+5 x+6=(x+2)(x+3)\) \(x^{2}+7 x+10\) We are looking for two numbers that multiply to give 10 and add to give 7. The numbers 2 and 5 satisfy both conditions. Thus \(x^{2}+7 x+10=(x+2)(x+5)\) \(x^{2}+9 x+14\) We are looking for two numbers that multiply to give 14 and add to give 9. The numbers 2 and 7 satisfy both conditions. Thus \(x^{2}+9 x+14=(x+2)(x+7)\) Exercises: Factorise: \(x^{2}+9 x+20\) We are looking for two numbers that multiply to give 20 and add to give 9. The numbers 4 and 5 satisfy both conditions. Thus \(x^{2}+9 x+20=(x+4)(x+5)\) \(x^{2}+12 x+32\) We are looking for two numbers that multiply to give 32 and add to give 12. The numbers 4 and 8 satisfy both conditions. Thus \(x^{2}+12 x+32=(x+4)(x+8)\) \(x^{2}+20 x+75\) We are looking for two numbers that multiply to give 75 and add to give 20. The numbers 5 and 15 satisfy both conditions. Thus \(x^{2}+20 x+75=(x+5)(x+15)\) \(x^{2}+15 x+56\) We are looking for two numbers that multiply to give 56 and add to give 15. The numbers 7 and 8 satisfy both conditions. Thus \(x^{2}+15 x+56=(x+7)(x+8)\) \(x^{2}-5 x+6\) We are looking for two numbers that multiply to give 6 and add to give -5. The numbers -2 and -3 satisfy both conditions. Thus \(x^{2}-5 x+6=(x-2)(x-3)\) \(x^{2}-17 x+30\) We are looking for two numbers that multiply to give 30 and add to give -17. The numbers -15 and -2 satisfy both conditions. Thus \(x^{2}-17 x+30=(x-15)(x-2)\) \(x^{2}-9 x+14\) We are looking for two numbers that multiply to give 14 and add to give -9. The numbers -7 and -2 satisfy both conditions. Thus \(x^{2}-9 x+14=(x-7)(x-2)\) \(x^{2}-15 x+44\) We are looking for two numbers that multiply to give 44 and add to give -15. The numbers -11 and -4 satisfy both conditions. Thus \(x^{2}-15 x+44=(x-11)(x-4)\) \(x^{2}-18 x+80\) We are looking for two numbers that multiply to give 80 and add to give -18. The numbers -10 and -8 satisfy both conditions. Thus \(x^{2}-18 x+80=(x-10)(x-8)\) \(x^{2}-14 x+40\) We are looking for two numbers that multiply to give 40 and add to give -14. The numbers -10 and -4 satisfy both conditions. Thus \(x^{2}-14 x+40=(x-10)(x-4)\) \(x^{2}-30 x+56\) We are looking for two numbers that multiply to give 56 and add to give -30. The numbers -28 and -2 satisfy both conditions. Thus \(x^{2}-30 x+56=(x-28)(x-2)\) \(x^{2}+x-6\) We are looking for two numbers that multiply to give -6 and add to give 1. The numbers 3 and -2 satisfy both conditions. Thus \(x^{2}+x-6=(x-2)(x+3)\) \(x^{2}+x-30\) We are looking for two numbers that multiply to give -30 and add to give 1. The numbers 6 and -5 satisfy both conditions. Thus \(x^{2}+x-30=(x-5)(x+6)\) \(x^{2}-5 x-14\) We are looking for two numbers that multiply to give -14 and add to give -5. The numbers -7 and 2 satisfy both conditions. Thus \(x^{2}-5 x-14=(x+2)(x-7)\) \(x^{2}-7 x-44\) We are looking for two numbers that multiply to give -44 and add to give -7. The numbers -11 and 4 satisfy both conditions. Thus \(x^{2}-7 x-44=(x+4)(x-11)\) \(x^{2}+2 x-80\) We are looking for two numbers that multiply to give -80 and add to give 2. The numbers 10 and -8 satisfy both conditions. Thus \(x^{2}+2 x-80=(x-8)(x+10)\) \(x^{2}+3 x-40\) We are looking for two numbers that multiply to give -40 and add to give 3. The numbers 8 and -5 satisfy both conditions. Thus \(x^{2}+3 x-40=(x-5)(x+8)\) \(x^{2}+4 x-21\) We are looking for two numbers that multiply to give -21 and add to give 4. The numbers 7 and -3 satisfy both conditions. Thus \(x^{2}+4 x-21=(x-3)(x+7)\) \(x^{2}-x+56\) This quadratic cannot be factored over the integers because there are no two numbers that multiply to give 56 and add to give -1. \(x^{2}-3 x+2\) We are looking for two numbers that multiply to give 2 and add to give -3. The numbers -1 and -2 satisfy both conditions. Thus \(x^{2}-3 x+2=(x-1)(x-2)\) \(x^{2}-3 x-10\) We are looking for two numbers that multiply to give -10 and add to give -3. The numbers -5 and 2 satisfy both conditions. Thus \(x^{2}-3 x-10=(x+2)(x-5)\) \(x^{2}-5 x-14\) We are looking for two numbers that multiply to give -14 and add to give -5. The numbers -7 and 2 satisfy both conditions