Question

Find the solutions of x^3-6x-5. Is this possible without graphing?

Answer 1

no

Answer 2

if you plug in 1 for function you should get 0, which means
(x-1) is a factor

Answer 3

Yes

Answer 4

I say yes

Answer 5

it's possible, just use wolfram or similar software. Or factor it yourself...

Answer 6

http://www.algebrahelp.com/calculators/expression/oops/calc.do?expression=x^3-6x-5

see no solution

Answer 7

sure it is if you want to read a really really really long formula for roots third degree polynomials.

Answer 8

*

-1
so
it would be (x+1)

Answer 9

but a good hint is if it was made up by a math teacher, try x = 1 and x = -1 first

Answer 10

Well I started to type a reply and then seen all these people typing a reply at the same time lol

Answer 11

my calculator says it has 3 roots, and 1 is 1 of them

Answer 12

what imran said. try -1, say "oh wow i am shocked, it works!" then factor out (x+1)

Answer 13

it would be uneven

Answer 14

There's like twelve people here!

Answer 15

That is the only rational root, could then use synthetic substitution to get the depressed equation of degree 2, then solve that

Answer 16

(x+1)(x^2-x-5) is it factored

Answer 17

\[x^3-6x-5=(x+1)(\text{something})\] and find the something. then quadratic formuala

Answer 18

yeah do what mandolino said and you get the other 2 roots but they have \[\sqrt{reallyhardnumber - mumbojumbo}\] in them

Answer 19

ya you could also use the quadratic formula...to assure if it is correct

Answer 20

After synthetic with the root x=-1 you get x^2-x-5=0

Answer 21

you can use synthetic division if you like, but you can also just think what it has to be

Answer 22

Is there a better way to figure out that -1 is a root than guess-and-check?

Answer 23

In this case, yes I agree

Answer 24

x^2-x-5=0 has 2 real roots

Answer 25

just like when you divide regular numbers. no there is not a better way than guess and check. only possible rational zeros are
\[\pm1,\pm5\] so -1 is a good guess

Answer 26

this has to be the most answered question i have ever seen!

Answer 27

x = \[\frac{-(\sqrt{21}-1)}{2}\text{ or -1 or }\frac{\sqrt{21}+1}{2}\]

Answer 28

x = \[\frac{-(\sqrt{21}-1)}{2}\text{ or -1 or }\frac{\sqrt{21}+1}{2}\]

Answer 29

since when does the radical come first?

Answer 30

my calculator prints it like that

Answer 31

If you want to look it up in your book, satellite used what is called the rational roots (or zeros) theorem to get the list of possible rational roots.

Answer 32

lol i thought it might be technology

Answer 33

technology ftw

Answer 34

x^3-6x-5=x^3-x-5x-5
x(x^2-1)-5(x+1)
x(x-1)(x+1)-5(x+1)
(x+1)(x(x-1)-5)
(x+1)(x^2-x-5)
x=-1
-b+-sqrt(b^2-4ac)/2a
1+-sqrt(1+20)/2
1+-sqrt(21)/2
x=-1
x=1+sqrt21/2
x=1-sqrt(21)/2