how do you factor this equation? x^3+6x^2+12x+8 = 0

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how do you factor this equation? x^3+6x^2+12x+8 = 0

Mathematics
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with a ten foot pole ?
start by getting number that might work by factoring the last over the first 1,8,2,4 ------ makes life a little easier 1
factor by grouping

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since all the terms are + already we know there are no positive solutions to be had so lets go with - tests
lets try x = -2 -2 | 1 6 12 8 0 -2 -8 -8 ----------- 1 4 4 0 <- its a good one casue we got 0 i n the end
x=-2 is a solution; (x+2) (x^2+4x+4) the quad factors quite nicely as well to get: (x+2)(x+2)(x+2) = (x+2)^3
why did you use -2? not 2
also how did you get (x+2)
or: (x^3+6x^2)+(12x+8)=0 x^2(x+6)+4(3x+2)=0 (x^2+4)(x+6)(3x+2)=0 x=-6 x=-2/3 x^2=4 \[x=\sqrt{4} = \pm 2\] then plug check to see which one solution, I think x=-2 is solution
Algebra I style : )
thank you for your medal
the rule for wigns tells me that if all are + then there are no poistive roots to be had; so I went with the negative versions of my gene pool
rule for signs.... its like im having a stroke lol
have to set factors equal to zero and solve...so much easier doing this stuff than my own...grrhhhh, lol!
the way amis do is call long division,you can do that way too ,but is hard to explant
x = sqrt(4) = 2 NOT +-2
not long division; close tho; synthetic division ;)
square= +-
sqrt(4) = 2 2^2 = +- sqrt(4)
got it lol i read it wrong
yes ,almos close
now the question asked to factor it; not find roots right?
(x+2)^3 is the factored form then

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