How do you solve #x^2(4-x)(x+6)<0#?

1 Answer
Leland Adriano Alejandro
Jul 23, 2016

The inequality is TRUE for values of x:
#x < -6" "# OR #" "x>4#

Explanation:

Since by solving for the values of x for each factor, we are going to have values #x=-6# and #x=0# and #x=4#

The intervals are #(-oo, -6)# and #(-6, 0)# and #(0, 4)# and #(4, +oo)#

Let us use test points for each interval

For #(-oo, -6)# , let us use #-7#

For #(-6, 0)# , let us use #-2#

For #(0, 4)# , let us use #+1#

For #(4, +oo)# , let us use #+5#

Let us do each test

At #x=-7" "#the value#" " " "x^2(4-x)(x+6)<0" "#TRUE
At #x=-2" "#the value#" " " "x^2(4-x)(x+6)<0" "#FALSE
At #x=+1" "#the value#" " " "x^2(4-x)(x+6)<0" "#FALSE
At #x=+5" "#the value#" " " "x^2(4-x)(x+6)<0" "#TRUE

Conclusion:

The inequality is TRUE for the following intervals
#(-oo, -6)# and #(4, +oo)#

OR

The inequality is TRUE for values of x:
#x < -6# OR #x>4#

God bless....I hope the explanation is useful.

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