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

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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