Need help on math problem?

Question

If y=6x^3-3x^2-2x+1, for what values of x is y > 0?

Answer 1

... since nobody answers this yet... i'll try....

6x³ - 3x² -2x + 1 > 0

to factor that, we can use grouping:

3x²(2x - 1) - (2x - 1) > 0

(2x - 1)(3x² - 1) > 0

There are 2 cases. One where both are positive or both are negatives.

I) 2x - 1 > 0 and 3x² - 1 > 0

2x > 1 ...... 3x² > 1

x > 1/2 ...... x² > 1/3

................. x > sqrt(1/3) or x < -sqrt(1/3)

Notice we have x > 1/2 and x > sqrt(1/3) or x < - sqrt(1/3), you need to think about the number line to see which range satisfy both.

It is x > sqrt(1/3)

II) 2x - 1< 0 and 3x² - 1 < 0

2x < 1..............3x² < 1

x < 1/2 ............. x²  < 1/3

........................ - sqrt(1/3) < x < sqrt(1/3)

So we have x < 1/2, and -sqrt(1/3) < x < sqrt(1/3), which range will satisfy both? -sqrt(1/3) < x < 1/2

Does that make sense? Is it what you are leaning?