To solve this question, we need to look at where this expression's value becomes zero. We can best do this through factoring the expression, seen here:

`x^4 - 7x^2 - 144 > 0`

We can actually factor this by trying to find two numbers that multiply to -144 and that...

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To solve this question, we need to look at where this expression's value becomes zero. We can best do this through factoring the expression, seen here:

`x^4 - 7x^2 - 144 > 0`

We can actually factor this by trying to find two numbers that multiply to -144 and that add to -7. After considering the factors of 144, it's clear that the numbers will be -16 and +9, giving us the following result:

`(x^2 - 16)(x^2+9) > 0`

The `x^2+9` term is never going to reach zero or be negative unless we are allowed to use imaginary numbers. The other term can be factored in the following way considering that it is the difference of two squares:

`x^2-16 = (x-4)(x+4)`

So, we are really considering where the following expression will be greater than zero:

`(x+4)(x-4) > 0`

This gives us three intervals to consider:

1) `x<-4`

2) `-4<x<4`

3) `x>4`

So to check each case, we simply choose a number inside of each interval to see whether the expression is positive or negative.

1) `x<-4`

For this case, let's use `x = -100`, just to make things easy. If `x = -100`, then the above expression becomes:

`(-100+4)(-100-4) > 0`

`-104*-96 > 0`

The great part here is that we don't even have to evaluate the expression. All we need to know is that two negative numbers multiplied together gives a positive result! Therefore, case 1 is valid, and the interval is part of our solution.

2) `-4<x<4`

For this case, let's choose `x = 0`, giving us:

`(0+4)(0-4) > 0`

`4*-4 > 0`

This expression is clearly false because a positive number multiplied by a negative number will always be negative! Therefore case 2 is false, and this interval is not part of our solution.

3) `x > 4`

Now, let's let `x = 100`. This gives the expression:

`(100+4)(100-4) > 0`

`104*96 > 0`

This result must be greater than zero because we are multiplying two positive numbers. So, case 3 is valid, and the interval is part of our solution.

So, our tests give us the following intervals for a solution:

`x in (-oo, -4) or x in (4, oo)`

We can check the above solution by graphing the original solution and examining for where the graph is above the y-axis.

Looks like we have the correct answer! Notice that our intervals will be open because our expression must be strictly less than 0 instead of less than or equal to zero.

I hope that helps!