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to factor this you need to find 2 numbers a and b whose product us 12 and whose sum is 8 the factors will be (x + a)(x + b)
x^2-8x+12=0 when you factor you need to find two numbers when multiblied must equal 12 and when added must equal -8. so the factors of 12 are: 1 and 12 2 and 6 3 and 4 -1 and -12 -2 and -6 -3 and -4 the pair that will give you -8 is -2 and -6 so (x-8)(x-2)=0 for the next one is ac method you have to multiply a and c which in this case is 4*9 and get 36 then you need to list factors of 36 and those factors are 1,36 2,18 3,12 4,9 5 doesnt go into 36 so it is not factor 6,6 and there is no more positive facors there are negative ones so they are -1,-36 -2,-18 -3,-12, -4,-9 and -6 and -6 those are all factors of 36. so you will always chose one pair wich will give you the sum of the midle term in this case the sum must be -12 so the numbers are -6 and -6 so then we will go and replace the midle term(-12x) with those two factors so we will have 4x^2-6x-6x+9=0 (note we did only replace the -12xby -6x and -6x because -12x= - 6x - 6x) so now we have 4 terms instead of 3 tearms and we will factor by grouping so we will have 4x^2-6x - 6x+9 =0 2x(2x-3) - 3(2x-3)=0 then since here both terms have commen factor which is (2x-3) we can factor it out and have (2x-3) (2x-3)=0 and that will be completing factoring