# Thread: Solving a polynomial equation

1. ## Solving a polynomial equation

I feel like this should be so easy, but I am stumped. I have to solve for y^3 = y.

This is how I am doing, but do not think it is correct:

y^3 = y
-y -y

y^3 - y = 0

y (y^2 - y) = 0

y^2 - y = 0
+y +y

y^2 = y
y y

y = 1

Solution:
y = 0
y = 1

Any help would be appreciated. Thank you!

2. ## Re: Solving a polynomial equation

$y^3 - y = 0$

$y(y^2-1) = 0$

$y(y-1)(y+1) = 0$

3. ## Re: Solving a polynomial equation

Originally Posted by tac
...
y^3 - y = 0

y (y^2 - y) = 0
...
That was a mistake in factoring, which is why you "lost" the solution y=-1. Skeeter's post shows the correct factorization.

Also, when you solve for zeros of a polynomial, it's best to do the entire factorization (with 0 on the right hand side) first, and then read off the roots (or use quadratic equations to get the roots of a quadratic that didn't factor nicely).
Picking off roots by dividing and manipulating as you go - the way you were working the problem - is not a good approach for many reasons. Again, see Skeeter's post for the correct way to do this.

4. ## Re: Solving a polynomial equation

Thank you!

Those perfect squares get me every time.

After factoring out the perfect square I am getting:

y = 0
y = 1
y = -1

Can there be three answers? When I plug in each one they all work.

5. ## Re: Solving a polynomial equation

Yes! Degree 3 means 3 solutions are possible.
There could be less than 3 answers, but NEVER more (well, except for the degenerate polynomial 0).
Degree n means n solutions are possible, but never more. There are never n+1 different roots for an n-th degree polynomial. This is counting complex roots.

Ex:
$y^2-3y+2 = 0$ has two solutions (y = 1, y=2).
$y^2+1 = 0$ has no real solutions, but two complex solutions (i and -i).
$y^2 = 0$ has only one solution (real or complex), which is y = 0.
Likewise
$y^2 -2y + 1 = 0$ has only one solutions (real or complex), which is y = 1.