Let us first factorize #f(x)=x^4-10x^3+27x^2-2x-40#. As #-1# is a zero of the function as

#f(0)=(-1)^4-10(-1)^3+27(-1)^2-2(-1)-40=1+10+27+2-40=0#, #(x+1)# is a factor of #f(x)#.

Also #f(2)=2^4-10*2^3+27*2^2-2*2-40=16-80+108-4-40=0#, hence #(x-2)# to is a factor of #f(x)#.

Dividing #f(x))=x^4-10x^3+27x^2-2x-40# by #(x+1)# and then by #(x-2)#, we get #x^2-9x+20# which can be further factorized to #(x-4)(x-5)# ad hence, the inequality we have is

#(x+1)(x-2)(x-4)(x-5)>=0#

Now, these zeros divide real number line in five parts

**(1)** #x<-1# - Here all the terms are negative, hence #f(x)# is positive. This forms part of solution.

**(2)** #-1 < x < 2# - Here while first term #(x+1)# is positive, all other terms are negative and hence #f(x)# is negative. Hence, this does not form part of solution.

**(3)** #2 < x < 4# - Here while first two terms #(x+1)# and #(x-2)# are positive, other two terms are negative and hence #f(x)# is positive. Hence, this forms part of solution.

**(4)** #4 < x < 5# - Here while first three term are positive, the last term #(x-5)# is negative and hence #f(x)# is negative. Hence, this does not form part of solution.

**(5)** #5 < x# - Here all the terms are positive and hence #f(x)# is positive. Hence, this forms part of solution.

Hence solution set for #x^4-10x^3+27x^2-2x-40>=# is #x<=-1# and #2 <= x <= 4# and #5 < x#.

graph{x^4-10x^3+27x^2-2x-40 [-10, 10, -40, 40]}