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

##### 1 Answer

#### Answer:

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

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

Hence

#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

#1-5+9-5 = 0#

Hence

#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

#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#