Question

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

Answer 1

c

Explanation:

source: trust me bro

Answer 2

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


`(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


`(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`