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

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Homework Help Overview

The discussion revolves around solving a polynomial equation, specifically 5(x+1)^3 = -5. The original poster expresses confusion regarding the solution provided by a tutor, which was x=0, and notes that this answer is incorrect according to their textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss dividing both sides by 5 and the challenges of finding solutions when the polynomial is set to -1. There are mentions of factoring and identifying roots, including both real and complex solutions.

Discussion Status

Some participants have offered guidance on how to approach the problem, suggesting methods such as substituting variables and factoring. There is acknowledgment of the complexity of the roots, particularly the imaginary ones, and the discussion reflects a mix of interpretations regarding the correct approach to the problem.

Contextual Notes

There is a reference to the original poster's textbook providing different answers, which raises questions about the assumptions made in the problem-solving process. The discussion also hints at the original tutor's potential oversight in verifying the solution.

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
[tex]5(x+1)^3=-5[/tex]
 
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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 [tex]\frac {-1\pm\sqrt{3i}} {2}[/tex] i don't know how to get this answer can some1 please help me immediately!:cry:
 
Last edited:
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 [tex]\frac {-1\pm\sqrt{3i}} {2}[/tex] 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.
 
Last edited:
thank u, i got it...thnx for the help
 

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