Find all values of x in this equation 3^x(x^3-2x^2-x+2)/x^3+6x^2+9x is less than or e

[appplejack]

Homework Statement

Find all values of x in this equation 3^x(x^3-2x^2-x+2)/x^3+6x^2+9x is less than or equal to zero. (how can I write it so that it would look good and not messy like above?)

Homework Equations

The Attempt at a Solution

simplify the equation so that I get the following.

3^x[x(x-1)^2+2]/x(x+3)^2 is less than or equal to zero.
(x cannot be 0 or 3) I'm stuck here and helpless from this point

 

[Mark44]

Homework Statement

Find all values of x in this equation 3^x(x^3-2x^2-x+2)/x^3+6x^2+9x is less than or equal to zero. (how can I write it so that it would look good and not messy like above?)

Homework Equations

The Attempt at a Solution

simplify the equation so that I get the following.

3^x[x(x-1)^2+2]/x(x+3)^2 is less than or equal to zero.
(x cannot be 0 or 3) I'm stuck here and helpless from this point

Here's your inequality. If you Quote my post, you can see the LaTeX that I used.
$$3^x\frac{x^3 - 2x^2 -x + 2}{x^3 + 6x^2 + 9x} \leq 0$$

You factored the denomimator, and that work looks fine. You also need to factor the numerator polynomial. What you have is not a factorization. The only possible rational roots of the numerator are +1 and -1, and +2 and -2. This means that x - 1, x + 1, x - 2, or x + 2, respectively, are potential factors.

Once you get things factored, we can go from there.

 

[LCKurtz]

Homework Statement

Find all values of x in this equation 3^x(x^3-2x^2-x+2)/x^3+6x^2+9x is less than or equal to zero. (how can I write it so that it would look good and not messy like above?)

The title says less or equal to e and above says less or equal 0. Which is it? Also it isn't clear what is in the denominator. You can use the X² button for superscripts. Also you can read here about using LaTeX:

https://www.physicsforums.com/showthread.php?t=386951

[Edit] I see now that the title of your post was truncated. All the more reason not to put equations in the title!

 

[epenguin]

You say you're stuck, but we don't quite know whether you're stuck just on a mechanical part, the factorisation of the numerator polynomial, or whether overall you don't know where you're going.

For the first, it is not that difficult and I think it's a pity to be stuck on a mechanical part so I think it's OK to suggest: look at the first two terms together, look at the last two terms together, suggest anything to you?

For where you're going, you've got some factors multiplied together and divided by some other factors multiplied together. Without even looking at the factors, if they were just A, B, C,... in general what conditions on the A, B, ... in something like ABCD/EFGH give you that the result is >/= 0 . BTW notice you already have a perfect square for one of the factors.

 

[HallsofIvy]

Assuming you meant "less than or equal to 0", the first thing I would do is try to factor those cubics. The "rational root theorem" says that the only possible rational roots (the only kind that gives simple factors) must be integers (since the leading coefficients are both 1) that evenly divide the constant term.

The constant term for $x^3-2x^2-x+2$ is 2 so the only possible rational roots are $\pm 1$, $\pm 2$. Try those.

As for the denominator, $x^3+6x^2+9x= x(x^2+ 6x+ 9)$ and the quadratic is a perfect square.