How do you find the roots of #x^3-6x^2+13x-10=0#?

1 Answer
Sep 15, 2016

Answer:

#x=2#

Explanation:

#x^3-6x^2+13x-10=0#

#x^3-3(x)^2(2)+3(2)^2x+x-2^3-2=0#

#(x^3-3(x)^2(2)+3x(2)^2-2^3)+x-2=0#

We can factorize using the polynomial identity that follows:
#(a-b)^3= a^3-3a^2b+3ab^2+b^3#

where in our case #a=x# and #b=2#

So,
#(x-2)^3+(x-2)=0# taking #x-2# as common factor
#(x-2)((x-2)^2+1)=0#
#(x-2)(x^2-4x+4+1)=0#
#(x-2)(x^2-4x+5)=0#

#x-2=0# then #x=2#
Or
#x^2-4x+5=0#
#delta=(-4)^2-4(1)(5)=16-20=-4<0#
#delta <0rArr# no root in R