# Solve the inequality: x^4 + 5x^3 - 16x^2 - 80x < 0

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### 2 Answers

Another approach again starts with factoring the polynomial:

`x(x+5)(x-4)(x+4)<0` . The function on the left is 0 at -5,-4,0,4.

Since all of the zeros have multiplicity 1, the function changes sign at the zeros. We now test values of the function on the intervals created by the zeros:

On `(-oo,-5)` we find `f(-6)=120>0`

So the function is positive on the entire interval.

**On `(-5,-4)` we find `f(-4.5)=-9.5625<0` so the function is less than zero on this entire interval.**

On `(-4,0)` we find `f(-1)=60` so the function is positive on the whole interval.

**On `(0,4)` we find `f(1)=-90<0` so the function is negative on this interval.**

On `(4,oo)` we find `f(5)=450>0` so the function is positive on this interval.

**Thus the intervals where the function is negative are `(-5,-4)` and `(0,4)` **

The inequality to be solved is x^4+5x^3-16x^2-80x<0

x^4 + 5x^3 - 16x^2 - 80x < 0

=> x^3(x + 5) - 16x(x + 5) < 0

=> (x^3 - 16x)(x + 5) < 0

=> x(x + 5)(x - 4)(x + 4) < 0

The product of the factors is negative if either only 1 of them is negative of if three of them are negative. Consider the following cases:

x <0, (x + 5)> 0, (x - 4) > 0 and (x + 4) > 0

This is not possible as x has to be less than 0 and more than 4 at the same time

x > 0, (x + 5)< 0, (x - 4) > 0 and (x + 4) > 0

This is not possible as x has to be greater than 0 and less than -5 at the same time

x > 0, (x + 5) > 0, (x - 4) < 0 and (x + 4) > 0

This is possible when as x > 0, x > -5 , x < 4 and x > -4 for x in (0, 4)

x > 0, (x + 5)> 0, (x - 4) > 0 and (x + 4) < 0

This is not possible as x has to be greater than 0 and less than -4 at the same time.

Again, for 3 factor being less than 0 and one greater than 0

x > 0, (x + 5)< 0, (x - 4) < 0 and (x + 4) < 0

This is not possible as x has to be more than 0 and more than -5 at the same time

x < 0, (x + 5) > 0, (x - 4) < 0 and (x + 4) < 0

This is possible for values of x that lie in (-5, -4)

x < 0, (x + 5) < 0, (x - 4) > 0 and (x + 4) < 0

This is not possible as x has to be less than 0 and more than 4 at the same time

x < 0, (x + 5) < 0, (x - 4) < 0 and (x + 4) > 0

This is not possible as x has to be less than -5 and more than -4 at the same time.

**The solution of the inequality is `(-5, -4)U(0, 4)` **