`(x+2)(4x^2 - 9) = 0`

`(x+2)(2x + 3)(2x - 3) = 0`

`x+2 = 0 and 2x + 3 = 0 and 2x - 3 = 0`

`x = -2 and x = -3/2 and x = 3/2`

**x = -2, -3/2, and 3/2**

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`(x+2)(4x^2 - 9) = 0`

`(x+2)(2x + 3)(2x - 3) = 0`

`x+2 = 0 and 2x + 3 = 0 and 2x - 3 = 0`

`x = -2 and x = -3/2 and x = 3/2`

**x = -2, -3/2, and 3/2**

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`4x^3+8x^2-9x-18=0`

Factor the greatest common factor from each group.

`4x^2(x+2)-9(x+2)=0`

Since `x+2` is a common factor, then

`(4x^2-9)(x+2) =0`

Now factor the binomial by using the difference of squares formula because both terms are perfect squares. The difference of squares formula is `a^2-b^2 = (a-b)(a+b)`

`(2x-3) (2x+3) (x+2)=0`

Set each of these factors equal to 0.

`2x-3=0`

`2x+3=0`

`x+2=0`

Solving for the first equation:

`2x-3=0`

Add `3` to both sides.

`2x=3`

Divide both sides by `2`

`x=3/2`

Next equation.

`2x+3=0`

Subtract `3` from both sides.

`2x=-3`

Divide both sides by `2`

`x=-3/2`

Last equation

`x+2=0`

Subtract `2` from both sides.

`x=-2`

**The complete solution set is:**

`x= 3/2, -3/2, -2`