solve the inequality for x x^3-5x greater then or equal to 4x^2 i simplified to x(x+5)(x-1) but its saying that is incorrect

\[x^{3}-5x \ge 4x ^{2}\]

I'm solving hold on

i figured it out and its \[x(x-5)(x+1)\] how do i figure out which are included in the solution set

\[x^3-5x\geq 4x^2\] \[x^3-4x^2-5x\geq0\] \[x(x^2-4x-5)\geq 0\] \[x(x+4)(x-5)\geq0\] zeros are at -4, 0 and 5and since this is a cubic polynomial with positive leading coefficient, it will be negative, then positive, then negative, then positive (think of the picture) you want it to be positive, so answer is \[[-4,0]\cup [5,\infty)\]

in your head draw this picture |dw:1314549885298:dw|

it says the answer is (-1,0] U [5, infinity)

because if you factor out (x+4)(x-5) you get \[x ^{2}-1x-30\] which isnt what i want

solution x>=5 -1<=x<=0

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