Question

What are the roots of `x^4-6x^3+14x^2-14x+5=0` with their multiplicities?

Answer 1

The roots are:

`x=1` with multiplicity `2`

`x=2+-i` each with multiplicity `1`

Explanation:

Given:

`x^4-6x^3+14x^2-14x+5=0`

Note that the sum of the coefficients is `0`, that is:

`1-6+14-14+5 = 0`

Hence `x=1` is a root and `(x-1)` a factor:

`x^4-6x^3+14x^2-14x+5 = (x-1)(x^3-5x^2+9x-5)`

Note that the sum of the coefficients of the remaining cubic is also `0`, that is:

`1-5+9-5 = 0`

Hence `x=1` is a root again and `(x-1)` a factor again:

`x^3-5x^2+9x-5 = (x-1)(x^2-4x+5)`

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(x-2)` and `b=i` as follows:

`x^2-4x+5 = x^2-4x+4+1`

`color(white)(x^2-4x+5) = (x-2)^2-i^2`

`color(white)(x^2-4x+5) = ((x-2)-i)((x-2)+i)`

`color(white)(x^2-4x+5) = (x-2-i)(x-2+i)`

So the remaining two zeros are:

`x = 2+-i`