(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.

The inequality x^3-24>-5x^2+2x has to be solved.

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

First factor the expression.

x^3 - 2x^2 + 7x^2 - 14x + 12x - 24 > 0

x^2(x - 2) + 7x(x - 2) + 12(x - 2) > 0

(x^2 + 7x + 12)(x - 2) > 0

(x^2 + 3x + 4x + 12)(x -2) > 0

(x(x + 3) + 4(x + 3))(x - 2) > 0

(x + 4)(x + 3)(x - 2) > 0

This is true when all the factors are positive or if two of the factors are negative and the third is positive.

This gives the following cases:

- (x + 4) > 0 , (x + 3)> 0, (x - 2) > 0

x > -4, x > -3, x > 2

This is true when x > 2

- (x + 4) > 0 , (x + 3)< 0, (x - 2) < 0

x > -4, x < -3, x < 2

This is true when x lies in (-4, -3)

- (x + 4) < 0 , (x + 3)> 0, (x - 2) < 0

x < -4, x > -3, x < 2

This is not true for any value of x.

- (x + 4) < 0 , (x + 3)< 0, (x - 2) > 0

x < -4, x < -3, x > 2

This is not true for any value of x

The solution of the inequality is `(-4, -3)U(2, oo)`