Also, since the solution intervals have already been given, I'll just add that the answer to the question you actually asked is **yes.** It's good to know that for *any cubic expression, *there will always be at least one interval where the expression is positive and at least one interval where it is negative. As long as `x ` is "positive enough", the expression will be positive, and if `x ` is "negative enough", the expression will be negative. Of course, what constitutes positive or negative enough depends on the cubic; `x^3 ` is positive as long as `x>0, ` but `x^3-10^(30) ` is not positive until after `x ` equals `10,000,000,000.`

First, find the zeros for the left side, aka solve x^3 - 13x + 12 = 0. It factors into (x+4)(x-1)(x-3). So, we have x+4 = 0, x-3 = 0, and x-1 = 0. So, x = -4, 3, 1. Then, you plot these on a number line, like in the image. They break up the number line into 4 different sections. You pick one number in each section, like I showed in the image, and plug it into the left side of the inequality.

For x = -5, x^3 - 13x + 12 = -48, a negative number, meaning all x's in this section would result in a negative number.

For x = 0, x^3 - 13x + 12 = 12, a positive number. So, this entire section would be positive.

For x = 2, x^3 - 13x + 12 = -6, a negative number. So, this entire section would be negative.

For x = 4, x^3 - 13x + 12 = 24, a positive number. So, this entire section would be positive.

Now, going back to the inequality, we are looking for all the x's would give a positive value (. . .> 0). That would be sections 2 and 4, intervals. Also, these intervals would be parenthesis instead of brackets since the inequality doesn't include "is equal to". So, the solution set is:

(-4,1) U (3,inf)

The solution of the inequality `x^3 - 13x + 12 > 0` has to be determined.

First, factor the polynomial.

x^3 - 13x + 12

= x^3 - x^2 + x^2 - x - 12x + 12

= x^2(x - 1) + x(x - 1) - 12(x - 1)

= (x - 1)(x^2 + x - 12)

= (x - 1)(x^2 + 4x - 3x - 12)

= (x - 1)(x(x + 4) - 3(x + 4))

= (x - 1)(x - 3)(x + 4)

(x - 1)(x - 3)(x + 4) > 0 in the following cases:

- x - 1 > 0, x - 3 < 0, x + 4 < 0

=> x > 1 , x < 3, x < -4

This has no solution.

- x - 1 < 0, x - 3 > 0, x + 4 < 0

=> x < -1 , x > 3, x < -4

This again has no solution

- x - 1 < 0, x - 3 < 0, x + 4 > 0

x < 1, x < 3 and x > -4

This is true for `x in (-4, 1)`

- x - 1 > 0, x - 3 > 0 and x + 4 > 0

=> x > 1 , x > 3 and x > -4

This is true for `x in (3, oo)`

**The solution of the inequality x^3 - 13x + 12 > 0 is `x in (-4, 1)U(3, oo)` **

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