x^2+8x+12

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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

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