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 $\displaystyle x = 3$.
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
$\displaystyle x=-1,\ g(x)=\frac{5}{5}=(-1)^2$