Solve for x in the polynomial

1. Oct 20, 2005

aisha

Hi i need some help solving this polynomail a tutor helped me with this question getting x=0 but this answer was not correct can someone else please help me out?

The question is
$$5(x+1)^3=-5$$

2. Oct 20, 2005

quasar987

Mmmh, divide by 5 on each side.

3. Oct 20, 2005

aisha

I already did it that way and then solved for x but that answer was wrong. x does not equal 0 in my text book it says the answers are -2 and $$\frac {-1\pm\sqrt{3i}} {2}$$ i dont know how to get this answer can some1 please help me immediately!!!!!

Last edited: Oct 20, 2005
4. Oct 20, 2005

quasar987

I tought you were only looking for the real part. I can't help you with the imaginary roots, sorry.

5. Oct 20, 2005

Diane_

Start by dividing by 5.

The problem is that it's not easy to find solutions to a polynomial when it's equal to, say, -1, but it's relatively very easy to find roots when the polynomial equals 0. So - multiply out the (x + 1)^3, move the -1 over to the left-hand side, and start factoring. Since you know what the roots are already, it should be easy to see that you'll have one linear factor and one quadratic one, so factoring the resulting cubic should be a snap.

6. Oct 20, 2005

ivybond

Aisha - how did your tutor get x=0 ?
Could not s/he plug 0 in original equation and see that it does not work?
Maybe it's time to find a more experienced tutor. :uhh:
(x+1)3 = -1
If you want you can substitute another variable (say, t) for (x+1).
t3 = -1
What's a real number "t" that's a cubic root of -1?
What's x now?
How about finding remaining complex roots?
Another way (as Diane_ suggested)
but do it for "t" instead, using formula
a3 + b3 = ?
After you cross all "t"s go back to
t = x +1, and determine corresponding "x"s with and/or without dotted "i"s.

Last edited: Oct 21, 2005
7. Oct 21, 2005

aisha

thank u, i got it...thnx for the help