# Thread: polynomial of degree 3

1. ## polynomial of degree 3

Hi,
how would i find the roots for the following:

$\displaystyle x^3-4x+x+6=0$

thanks.

2. Originally Posted by dragonation
Hi,
how would i find the roots for the following:

$\displaystyle x^3-4x+x+6=0$

thanks.
I assume that this is actually

$\displaystyle x^3 - 4x^2 + x + 6 = 0$.

Using the remainder and factor theorems, we can see that when $\displaystyle x= -1$, the function $\displaystyle = 0$.

So $\displaystyle x + 1$ is a factor.

Using long division, we find that

$\displaystyle x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6)$

$\displaystyle = (x + 1)(x - 2)(x - 3)$.

So now using the Null Factor Law:

$\displaystyle (x + 1)(x - 2)(x - 3) = 0$

$\displaystyle x = -1$ or $\displaystyle x = 2$ or $\displaystyle x = 3$.

3. An unconventional way to view it is as follows,
given that there are no more than 3 solutions,
and works fine if the solutions are integers.

$\displaystyle x^3-4x^2+x+6=0$

$\displaystyle x^3-4x^2=-x-6$

$\displaystyle x^2(x-4)=-x-6$

$\displaystyle x^2(4-x)=x+6$

$\displaystyle x^2=\frac{x+6}{4-x}=g(x)$

You can examine x quite easily then...

$\displaystyle x=0,\ g(x)=\frac{6}{4}$

$\displaystyle x=1,\ g(x)=\frac{7}{3}$

$\displaystyle x=2,\ g(x)=\frac{8}{2}=4=2^2$

x=2 is a solution

$\displaystyle x=3,\ g(x)=\frac{9}{1}=3^2$

x=3 is a solution

$\displaystyle x=5,\ g(x)=\frac{10}{0}$

and higher values of x cannot yield a positive solution

$\displaystyle x=-1,\ g(x)=\frac{5}{5}=(-1)^2$