How do you find all the zeros of #x^3-6x^2+13x-10# with its multiplicities?

1 Answer
Apr 17, 2016

Answer:

#2, 2+-i#

Explanation:

Explanation:

We will use the Rational Root Theorem:

If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.

So, the candidates for roots are:

#+-1, +-2,+-5,+-10#

We discover than 2 is a root.

Now we divide the polymorph by (x-2) to discover the other roots:

#(x^3-6x^2+13x-10)/(x-2)=x^2+(-4x^2+13x-10)/(x-2)#
#=x^2-4x+(5x-10)/(x-2)=x^2-4x+5#

This second degree polynom is solved with the quadratic formula:

#x=(4+-sqrt(4^2-4*1*5))/2=(4+-sqrt(16-20))/2=(4+-sqrt(-4))/2=(4+-2i)/2=2+-i#