Question

Hi everyone! How can i find the roots of x^3+x^2+x-1 ? i know that one of them is real, and the other two are imaginary, but how can i get them?

Answer 1

by rational root theorem you have list all the real roots possible.

here your constant term is 1 . The possible roots of 1 are + 1, -1

Your leading coefficient is 1. The possible roots of this are again, +1 and -1

now divide each root of constant term by each root of the leading coefficient.

So the possibility here are again + 1, -1

So substitute x = + 1 if you y turns 0 then x = + 1 is a root. So (x -1 ) will be a factor

Otherwise,

Substitute x = -1 in your function, if y turns zero then x = -1 is root . So (x + 1) will be a factor.

Once you get that use synthetic division of the polynomial by this one factor that you know to find the next factor.

Answer 2

This looks more like an expression rather than a equation. Is their a equal sign somewhere?

Answer 3

"f(x) = 0" is implied when "finding roots" :)

Answer 4

but here +1 and -1 does not work, then how do you find a fraction that will fit this function

Answer 5

Oh, well never mind.

Answer 6

All I can think of is just put it in the graphing calculator and be done wit hit

Answer 7

I can´t use the calculator or the pc; i know that the solutions are x=0.543689 x=-0.771845-1.11514i x=-0.771845+1.11514i but the thing is how to calculate them..

Answer 8

trial and error I guess

Answer 9

@amistre64 could you help pl?

Answer 10

x^3+x^2+x-1 ; let x = u-1/3

(u-1/3)^3 + (u-1/3)^2 + u - 1/3 - 1

u^3 + 2u/3 - 34/27 = 0

and then Cardano might be useful to us

Answer 11

but it might be helpful to know if youve come across Cardano before, if not then this is beyond the scope of your current material

Answer 12

I think the Cardano is for real roots; one idea i have is that if you know that it has 2 imaginary roots, then it must be (being presumptuous) of the form:\[(x-r)(a+bi-x)(a-bi-x)\]
\[x^3-(2a+r)x+(a^2+b^2+2ar)x-r(a^2+b^2)\]
i wonder if we can equate the given setup with this.

the real value from the wolf tho is horrendous, so im not even sure it is possible to find exact forms of these things by hand in a reasonable amount of time

Answer 13

why did you choose 1/3? Is it random?

Answer 14

its a reduction trick

\[P_n(x)=c_nx^n+c_{n-1}x^{n-1}+...\]
\[if~x=v-\frac{c_{n-1}}{n}, ~then ~P_n(v)=b_nv^n+0v^{n-1}+b_{n-2}v^{n-2}+...\]

Answer 15

SOLUTION!
Hi everyone! I have been looking for info and maybe this would be useful for someone who had the same problem as me:
the only real root isn´t integer or rational number, so I draw the graph and saw that there was one root between x=0 and x=1.
Then, I applied Newton´s Method (using derivatives to find closer numbers to the root) starting at X=1, obtaining (calculating the zero of the derivative of the function at x=1) the next X, which is X=2/3=0.6666... which is close to the root 0.543689... of the function.
I applied the method again, and the next X is X=165/297=0.555555... very close to the root! If you want more accuracy, you just have to repeat the method.

Then, if you want to calculate the imaginary roots, you just do Ruffini with the real root you obtained previously ( obviously, the more accurate you find the real root, the more precise imaginary roots you will obtain) and solve the quadratic.

Hope this will help if anyobody asked this before.

Answer 16

Newtons method is nice yes, but the question we had was more in regards to how exact, or accurate, you needed the roots to be in the first place.

Answer 17

Yeah, I know... anyway, thank you so much for the attention.
Best regards from Spain.