polynomial of degree 3

**dragonation** · March 5th 2010, 04:16 AM

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

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

thanks.

**Prove It** · March 5th 2010, 04:24 AM

Originally Posted by dragonation

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

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

thanks.

I assume that this is actually

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

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

So $x + 1$ is a factor.

Using long division, we find that

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

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

So now using the Null Factor Law:

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

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

**Archie Meade** · March 5th 2010, 06:28 AM

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.

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

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

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

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

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

You can examine x quite easily then...

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

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

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

x=2 is a solution

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

x=3 is a solution

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

and higher values of x cannot yield a positive solution

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

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