How can I solve for x in a polynomial equation with the help of a tutor?

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To solve the polynomial equation 5(x+1)^3 = -5, start by dividing both sides by 5, simplifying to (x+1)^3 = -1. The correct solutions are x = -2 and x = (-1 ± √3i)/2, indicating both real and complex roots. Factoring the cubic equation after moving -1 to the left side can help identify these roots. If a tutor provided an incorrect solution of x = 0, it may be beneficial to seek a more experienced tutor for guidance. Understanding both real and imaginary solutions is crucial for solving polynomial equations effectively.
aisha
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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
 
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Mmmh, divide by 5 on each side.
 
I already did it that way and then solved for x but that answer was wrong. x does not equal 0 in my textbook it says the answers are -2 and \frac {-1\pm\sqrt{3i}} {2} i don't know how to get this answer can some1 please help me immediately!:cry:
 
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I tought you were only looking for the real part. I can't help you with the imaginary roots, sorry.
 
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.
 
aisha said:
I already did it that way and then solved for x but that answer was wrong. x does not equal 0 in my textbook it says the answers are -2 and \frac {-1\pm\sqrt{3i}} {2} i don't know how to get this answer can some1 please help me immediately!:cry:
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. :rolleyes:
(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)
move the -1 over to the left-hand side, and start factoring
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.
 
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thank u, i got it...thnx for the help
 

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