solve the inequality and give solution in interval notation?

Question

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

Answer 1

Firstly, since its a +x^3 type graph, its going to go up, down, and up again.
Now, first solve it = 0. x^3 + 2x^2 - 3x = 0, so x(x^2+2x-3) = 0, so x(x+3)(x-1) = 0. That means x = -3, 0 or 1 are the three roots.

Thus, we know it starts off very small, goes up to 0 when x=-3, keeps going up into the positives and down again to x=0, down into the negatives and up again to x=1, and finally up to infinity for x>1.

Thus the parts where it is positive are between -3 and 0, and > 1.
Or (-3,0) union (1,∞)

Answer 2

Solve first for x intercepts (zeros).
x(x² + 2x - 3) = 0
x(x + 3)(x - 1) = 0
zeros at -3, 0, 1

Since the x^3 coefficient is >0, the graph goes low left and high right, squiggle in the middle. So the polynomial > 0 when
x in (-3,0) and (1,∞).

Answer 3

First off, we factor the left hand side of the inequality.

x(x^2 + 2x - 30) > 0

Are you sure you typed this question correctly? I'm going to assume the middle term has no 2 in it, giving us

x(x^2 + x - 30) > 0
x(x+6)(x-5) > 0

At this point, you note that your critical points are when the left hand side is equal to 0. That is, x = 0, -6, and 5

You then make a number line, with 0, -6, and 5 being the separators.


-------------------------------------
-6||||||||||||||||||||||0||||||||||||||||||||||||||5

At this point, you test each of the intervals into the given polynomial to see if you get a positive or negative number. You can even test 1,000,000,000, and you'll find that you get a positive number testing that. From that point, it will alternate.

Thus, since it's positive from 5 to infinity, it'll be negative from 0 to 5, positive from -6 to 0, and negative from -infinity to -6.

Since you're only interested in where it's strictly positive, your intervals are:

(-6,0) union (5 to infinity)

Answer 4

factor x

x^3+2x^2-3x>0 ==> x(x^2+2x-3)>0

then solve for x,

x( x -1)(x + 3) > 0

so x( x -1)(x + 3) = 0 has solution when x =0, x = 1, and x = -3

draw a table

x = - infinite ***-4 *** -3 ***-1*** 0 ***0.5*** 1 ***2***positive inf.
x = '''''''''-'''''''''''''' -''''''''' -''''''''''' -''''''''' 0''''''' +''''''''' +''''' + '''''''' +
x-1 =''''''' -''''''''''''' -''''' - ''''''''' - '''''''''' -'''''''' -''''''''' 0''''''+'''''''''''+
x+3 =''''''-'''''''''''''' -'''''''''' 0'''''''+ '''''''''+'''''''''+'''''''''' +''''' +''''''''' +
f(x) =''''' - ''''''''''' -'''''''''''''0 ''''''+ ''''''''' 0 ''''''' - ''''''' 0'''''+'''''''''+

Note f(x) = x(x-1)(x+3)

so f(x) > 0 when -3 < x <0 or 1< x