Question

x^2+8x+12

Answer 1

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)

Answer 2

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