## anonymous 5 years ago how do you factor this equation? x^3+6x^2+12x+8 = 0

1. amistre64

with a ten foot pole ?

2. amistre64

start by getting number that might work by factoring the last over the first 1,8,2,4 ------ makes life a little easier 1

3. anonymous

factor by grouping

4. amistre64

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

5. amistre64

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

6. amistre64

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

7. anonymous

why did you use -2? not 2

8. anonymous

also how did you get (x+2)

9. anonymous

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

10. anonymous

Algebra I style : )

11. anonymous

thank you for your medal

12. amistre64

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

13. amistre64

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

14. anonymous

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

15. anonymous

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

16. amistre64

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

17. amistre64

not long division; close tho; synthetic division ;)

18. anonymous

square= +-

19. amistre64

sqrt(4) = 2 2^2 = +- sqrt(4)

20. amistre64

got it lol i read it wrong

21. anonymous

yes ,almos close

22. amistre64

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

23. amistre64

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