# Solve for the roots of x^4+3x^3+x^2-3x-2.Show step by step process to explain the solution and answer

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It is not quite as easy to solve quartic equation (an equation with `x^4` in it) as it is to solve a quadratic because the equation to solve quartic equations is long and unwieldy, unlike the quadratic equations. The best way to solve this equation is to factorise it

Factorisation is the process of dividing the equation by linear factors to find the roots of the equation. Quartics typically split into 4 factors, so we are looking for 4 roots. Looking at the equation

`x^4+3x^3+x^2-3x-2=0`

I can see because of the -2 term at the end that x=0 cannot be a root, so I am going to try x=1

`(1)^4+3(1)^3+(1)^2-3(1)-2=1+3+1-3-2=0`

so x=1 is a root of the equation. This means that this equation has a factor at `(x-1)` because when you factorise something, the root is taken from x to give the factor.

We now divide the equation by the factor `(x-1)` and this is accomplished by using polinomial division. Polinomial division in this case is a process where you take the first two terms of the equation and see how many times your divisor will go into it. I will assume you know how to do polinomial division as its a long answer in itself but if you don't, I've put a link to quite an instructive page at the bottom.

When we divide `x^4+3x^3+x^2-3x-2=0` by `(x-1)` we get

`x^3+4x^2+5x+2=0`

This equation will now factor into 3 other equations giving us

`x^3+4x^2+5x+2=(x+?)(x+?)(x+?)`

If I try x=-1, I see that it also is a root of this equation

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

so (x+1) is also a root. If we again divide the equation by this we get:` `

`x^2+3x+2=0`

and this can be simply factorised into (x-1)(x-2).

Therefore `x^4+3x^3+x^2-3x-2=(x-1)(x+1)(x+1)(x+2)`

**which gives the roots as**

`x=1,x=-1,x=-2`

**Sources:**