#x^3-7x^2+5x=-1#
#x^3-7x^2+5x+1=0#
By inspection,
Let
#g(x)=x^3-7x^2+5x+1#
For #x=1#
#g(1)=(1)^3-7(1)^2+5(1)+1#
#=1-7xx1+5+1#
#=1-7+5+1#
#=0#
#g(1)=0#
1 is a root
#x-1#
is a factor
Now,
#g(x)=h(x)(x-1)#
#x^2(x-1)=x^3-x^2#
#x^3-7x^2+5x+1=x^3-x^2-6x^2+5x+1#
#x^3-7x^2+5x+1=x^2(x-1)-6x^2+5x+1#
#-6x(x-1)=-6x^2+6x#
#x^2(x-1)-6x^2+5x+1=x^2(x-1)-6x^2+6x-x+1#
#x^2(x-1)-6x^2+5x+1=x^2(x-1)-6x(x-1)-x+1#
#x^3-7x^2+5x+1=x^2(x-1)-6x(x-1)-x+1#
#-x+1=-1(x-1)#
#x^2(x-1)-6x(x-1)-x+1=x^2(x-1)-6x(x-1)-1(x-1)#
#x^3-7x^2+5x+1=x^2(x-1)-6x(x-1)-1(x-1)#
Rearranging
#x^3-7x^2+5x+1=(x-1)(x^2-6x-1)#
By the method of completing the squares
#x^2-6x-1=(x^2-6x+9)-(9+1)#
#x^2-6x+9=(x-3)^2#
#-(9+1)=-10#
#x^2-6x-1=(x-3)^2-10#
#x^2-6x-1=(x-3)^2-(sqrt10)^2#
#(a^2-b^2=(a-b)(a+b)#
#(x-3)^2-(sqrt10)^2=((x-3)-sqrt10)((x-3)+sqrt10)#
Simplifying
#x^2-6x-1=(x-(3+sqrt10))(x-(3-sqrt10))#
#(x-1)(x^2-6x-1)=#
#(x-1)(x-(3+sqrt10))(x-(3-sqrt10))#
Thus,
#x^3-7x^2+5x+1=#
#(x-1)(x-(3+sqrt10))(x-(3-sqrt10))#
Thus, f(x)=-1, at
#x=1, 3+sqrt10, 3-sqrt10#
Arranging in the Increasing order
#x=3-sqrt10,1,3+sqrt10#