How do you solve #x^4-10x^3+27x^2-2x-40>=0#?

2 Answers
Jul 21, 2016

Answer:

#p(x) ge 0# for #(x le -1) and (2 le x le 4) and (5 < x)#

Explanation:

Analyzing the constant term which is the product of all roots we get

#40=1cdot2^3 cdot 5#. Trying for #pm1,pm2,pm4,pm5# we get

#p(x)=x^4 - 10 x^3 + 27 x^2 - 2 x - 40#
#p(x)=(x+1)(x-2)(x-4)(x-5) ge 0#

Ordering the roots

#-1,2,4,5#

and knowing that for #x < -1->p(x)>0# we have by continuity

#-1 < x<2->p(x)<0#
#2 < x < 4 ->p(x) >0#
#4 < x < 5->p(x) < 0#
#5 < x ->p(x) > 0#

so

#p(x) ge 0# for #(x le -1) and (2 le x le 4) and (5 < x)#

Jul 21, 2016

Answer:

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

Explanation:

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]}