Question

How do you solve the polynomial `x^4-4x^3+5x^2-4x+4=0`?

Answer 1

Use the rational root theorem to help find one of the roots. Separate out the corresponding factor, then factor by grouping to find roots `2`, `2`, `i` and `-i`.

Explanation:

Let `f(x) = x^4-4x^3+5x^2-4x+4`

By the rational root theorem, any rational roots of `f(x) = 0` must be expressible in lowest terms as `p/q` for some integers `p`, `q`, where `p` is a divisor of the constant term `4` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational roots are:

`+-1`, `+-2`, `+-4`

Also since the signs of `f(x)` alternate for odd and even powers of `x`, `f(x) = 0` has no negative roots. So that only leaves the following possible rational roots:

`1`, `2`, `4`

Let's try each in turn:

`f(1) = 1-4+5-4+4 = 2`
`f(2) = 16-32+20-8+4 = 0`

So `x=2` is a root and `(x-2)` a factor.

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

The remaining cubic can be factored by grouping:

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

Putting it all together:

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

`= (x-2)^2(x-i)(x+i)`

So the `4` roots are `2`, `2`, `i` and `-i`

graph{x^4-4x^3+5x^2-4x+4 [-9.625, 10.375, -1.36, 8.64]}