Question 339897
x^3+2x^2-3x>0
first we find the zeros.
x(x^2+2x-3)=0
x(x+3)(x-1)=0
x=0, x=-3, x=1
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Now we test on either side of the zeros.
x=-4 y=-20
x=-2 y=6
x=1/2 y=-.875
x=2 y=10
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(-3, 0), (1, infinity) answer
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Ed
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{{{graph(500,500,-10,10,-10,10,x^3+2x^2-3x)}}}