# Thread: solving for x with 4th degree polynomial

1. ## solving for x with 4th degree polynomial

$5(x+1)^4-5$
the way i did it was
$5(x+1)^4=5$
$(x+1)^4=1$
$x+1 = 1$
$x = 0$
but, i know there is another answer of -2 because if you get $(-1)^4$ it is also 1...
so i think that i'm not doing this the proper way
can anyone show me the proper way to solve this?

2. Sorry, do you mean, solve $5(x+1)^4-5 = 0$ ?

EDIT:// I checked your work and there seems nothing wrong with your method. It is quite slick to think of $1^4 = (-1)^4 = 1$

You could always use polynomial expansion, foiling out $(x+1)^4$ if you like, but that wouldn't seem to be worth the effort

3. yes sorry!

4. Originally Posted by dorkymichelle
$5(x+1)^4-5{\color{red}=0}$
the way i did it was
$5(x+1)^4=5$
$(x+1)^4=1$
$x+1 = 1$
No, that's not quite correct, instead you have two possibilities, namely: $x+1 = {\color{red}\pm} 1$

5. My mistake; sorry

Failure: Plug in -1 and you get -5 on the LHS (for the original equation)
According to Failure,

$x+1=\pm 1$

$\therefore x +1 = 1$ OR $x+1= -1$

this gives:

$x = 1-1 = 0$ OR $x = -1-1 = -2$

Failure's(man, this name is weird!) method is correct. In fact, -1 is NOT one of the two values of x.

7. You are dead on with your explanation (that -2 is the other answer). I believe this is the step where the oversight was:
Originally Posted by dorkymichelle
$(x+1)^4=1$
$x+1 = 1$
When dealing with an equation involving powers, it is insufficient to take simply the principle root. For an equation of real numbers involving an even power, there are always two roots of positive numbers (the principal root and the negative of the principal root). So $x^8=2$ implies that $x=\sqrt[8]{2}$ or $x=-\sqrt[8]{2}$. For odd powers, there is only one root of a positive number. So $x^5=4$ means the only solution is $x=\sqrt[5]{4}$

If you are dealing with equations over the complex numbers, things are different. By the fundamental theorem of algebra, there $n$ roots for a polynomial of degree $n$. So $x^8=2$ has solutions $x=\sqrt[8]{2}e^{2\pi i k/8}$ for $k=0,1,\ldots,7$. So can picture the solutions as vertices of a polygon in the complex plane (as seen here), with one vertex at the principal root. For even powers you have the symmetry of flipping around the vertical axis which is not present in odd powers.

8. I'm sorry for double posting
however, on this post, i asked a question that dealt with the algebra part of the problem, it was an algebra concept that I couldn't understand, on the other post, it was a calculus concept.
Should I put it both in one thread next time?