Question

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

Answer 1

with a ten foot pole ?

Answer 2

start by getting number that might work by factoring the last over the first

1,8,2,4
------ makes life a little easier
1

Answer 3

factor by grouping

Answer 4

since all the terms are + already we know there are no positive solutions to be had so lets go with - tests

Answer 5

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

Answer 6

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

Answer 7

why did you use -2? not 2

Answer 8

also how did you get (x+2)

Answer 9

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

Answer 10

Algebra I style : )

Answer 11

thank you for your medal

Answer 12

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

Answer 13

rule for signs.... its like im having a stroke lol

Answer 14

have to set factors equal to zero and solve...so much easier doing this stuff than my own...grrhhhh, lol!

Answer 15

the way amis do is call long division,you can do that way too ,but is hard to explant

Answer 16

x = sqrt(4) = 2 NOT +-2

Answer 17

not long division; close tho; synthetic division ;)

Answer 18

square= +-

Answer 19

sqrt(4) = 2

2^2 = +- sqrt(4)

Answer 20

got it lol i read it wrong

Answer 21

yes ,almos close

Answer 22

now the question asked to factor it; not find roots right?

Answer 23

(x+2)^3 is the factored form then