Question

Solve the equation x^6+3x^5+x^4-5x^3-6x^2-2x=0 in the real number system

Answer 1

First, pull out a factor of $x$.

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

Notice that when $x=-1$ (which you can arrive at via the rational root theorem), you have

$(-1)^5+3(-1)^4+(-1)^3-5(-1)^2-6(-1)-2=-1+3-1-5+6-2=0$

which means you can pull out a factor of $x+1$. Upon dividing you get

$\dfrac{x^5+3x^4+x^3-5x^2-6x-2}{x+1}=x^4+2x^3-x^2-4x-2$

The rational root theorem will come in handy again, suggesting that $x=-1$ appears a second time as a root, which means

$\dfrac{x^4+2x^3-x^2-4x-2}{x+1}=x^3+x^2-2x-2$

Now this is more readily factored without having to resort to the rational root theorem. You have

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

so in fact, $x=-1$ shows up as a root for a third time.

So, you have

$x^6+3x^5+x^4-5x^3-6x^2-2x=x(x+1)^3(x^2-2)=0$

Two roots are obvious, $x=0$ and $x=-1$ (with multiplicity 3). The remaining two are given by

$x^2-2=0\implies x^2=2\implies x=\pm\sqrt2$