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)

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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)`