Solve the following polynomial inequality: x^3-24>-5x^2+2x

(We appreciate a 9th grade student tried and got the
answer. But he made a small error in the sign.)

To solve
x^3-24
>-5x^2+2x.

Solution:

Rearranging
we get:

x^3+5x^2-2x-24 >
0.

First we solve the equation:f(x) = x^3+5x^2-2x-24 =
 0.

Clearly f(2) = 2^3+5(2^2)-2*2-24 = 8+20-4-24 = 0. So
x-2 is a factor of  x^3+5x^2-2x-24.. So we can
write,

x^3+5x^2-2x-24 = (x-2)(x^2+ax+b).The leading term on
both sides is the same , that is, x^3. So, we can equate the constant terms and xterms
on both sides of this identity.

Constant terms -24 = -2b.
Or b = -24/(-2) = 12.

xterms: -2x = -2ax+bx =
-2ax+12x.Or

-2x-12x= 2ax. Or -14x = -2ax. Or a = -14x/(-2x)
= 7.

So x^2+ax+b =x^2+7x+12 =
(x+3)(x+4).

Therefore

x^3+5x^2-2x-24
> 0 implies  (x+4)(x+3)(x-2) = f(x) is  >
0.

When x > 2, all factors are positive. S o
x>2 is a valid solution.

When x between -3 and 2,
f(x) has (+)(+)(-) = -ve sign or f(x) <2. So -3<x<2 cannot be a
solution.

When -4<x<-3, f(x) has a sign
(+)(-)(-) is positive. Or f(x) > 0. So -4<x<3 is a valid
solution.

When x<4, f(x) has (-)(-)(-) = ve sign. Or
f(x) < 0. So x< -4 is not a valid
solution.

So x> 2 Or -4<x<-3 are the
valid solution for x. Or in interval notation,  ]-4,-3[ ,
 Or    ]2, infinity[ are the two open intervals. Open
implies end points are not included as the inequality f(x) > 0 indicates f(x) is
strictly greater than zero.