# Solve for the roots of x^4+3x^3+x^2-3x-2Show complete solution and explain the answer.

### 1 Answer | Add Yours

After checking for easy solutions such as completing the square, I resorted to long division. Although it takes several steps, it does work.

We know that x+1, x-1, x+2, and x-2 are the most likely roots, because they are the factors of -2. So we start with one of these.

I confess that I started with x-2 and it does not work. Next I tried x+2 as shown below.

`x^4 + 3x^3+x^2-3x -2` divided by `x+2`

`(x+2)(x^3+x^2-x-1)`

We need the `x^3` because that gives us `x^4` and `2x^3`

We need one more `x^3` that is why we have `x^2`

That gives us `3x^3` and `2x^2`

We only need one `x^2` so We need a `-x`

Now we have `-2x` and we need `-1x` , so we subtract one.

With the `-1` in the second factor, we know to try `x+1` or `x-1` next.

I chose `x+1` because I could see that it would give me a positive` ``x^2`` `

`(x+2)(x+1)(x^2-1)`

The last term is the difference of squares, so it is

`(x+2)(x+1)(x+1)(x-1)`

Setting each of the factors equal to 0 we get

**The roots are **` x=-2 `** , ` `**`x=-1`**` ` , ` `**`x=-1`**` ` , and `x=1` **

**Roots are -2, -1, and 1**