Solving For $X$ In A 4th Degree Polynomial

Question

The examples they use on my book are basically from basic arithmetic like 1, 2 and 3 to calculus and calculating derivatives, which is really annoying because I can't build the fundamental skill required to do harder questions. Anyhow, I have to solve an non-factorable inequality. Normally, I would use the quadratic formula but it's not quadratic, nor cubic.

The question is $x^4 + 2x^3 - 4x^2 - 6x \leq -3$. And I have to get an $x$ = something. I am actually really lost here because I can't even apply factor theorem in this case since no values work out.

I think this is a problem of my fundamentals though, please tell me how to solve things like these when they cannot be factored. Or we can just have a one on one discussion...

Any help is appreciated, thank you!

Answer 1

There is a quartic formula, but I do not recommend using it, since it is incredibly convoluted. Since you are trying to solve an inequality, it is probably fine to use a numerical approximation for the zeroes, which are about $.414, -1.732, 1.732,$ and $ -2.414$. I know from experience that $\sqrt{3} \approx 1.732$, and plugging in $\sqrt{3}$ and $-\sqrt{3}$ show that those are in fact roots. Dividing your polynomial by $x^2 - 3$ gives the polynomial $x^2 + 2x - 1$. Using the quadratic formula, we get that the other two roots are $-1 + \sqrt{2}$ and $-1 - \sqrt{2}$. If you are not allowed to use numerical approximations to find your roots, I would recommend plugging in square roots of small numbers (especially divisors of the constant term) in addition to small integers.

Answer 2

After manipulating a little bit and rearranging terms when the whole thing is $\leq 0$:

$$x^4 + 2x^3 - 4x^2 - 6x + 3 \\ x^4+2x^3-6x-4x^2+3 \\ x^4 + 2x(x^2-3)-4x^2+3 \\ x^4 + 2x(x^2-3)-3x^2-x^2+3 \\ x^4 + 2x(x^2-3)-3x^2-(x^2-3) \\ x^4 -3x^2 + 2x(x^2-3)-(x^2-3) \\ x^4 -3x^2 + (x^2-3)(2x-1) \\ x^2(x^2-3) + (x^2-3)(2x-1) \\ (x^2-3)(x^2+2x-1).$$

So your problem is now $(x^2-3)(x^2+2x-1) \leq 0$. So it appears that $x=\pm \sqrt{3}$ will bring about equality. If you use the quadratic formula, you will also find that $x=\pm \sqrt{2}-1$ are 2 more solutions. That means we've found all 4 roots(!).

After testing sample values within each of the intervals delimited by the 4 roots, you can show that the total valid domain is $x \in [-1-\sqrt{2}, -\sqrt{3}] \cup [\sqrt{2}-1,\sqrt{3}]$.