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

1 Answer
George C.
Mar 19, 2017

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#

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