I was helping a precalculus student with this question. The graph wasn’t given. My only idea was to find the inverse and try to find its domain. When trying to find the inverse, I arrived at $0=y^4+2y^3-15y^2+x$ and I was hoping to be able to factor it. However, besides using the quartic formula, I’m not […]

Question What is the minimum possible value of $a^{2}+b^{2}$ so that the polynomial $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least 1 root? Attempt I divided by $x^{2}$ and got $$x^{2}+\frac{1}{x^{2}}+2+a\left(x+\frac{1}{x}\right)+b-2=0$$ by letting $$x+\frac{1}{x}=X$$ the equation becomes: $$X^{2}+aX+(b-2)=0$$ $$\therefore X=\frac{-a\pm \sqrt{a^{2}-4b+8}}{2}$$ but I am not sure how to continue. If the polynomial has 1 root doesn’t that is should […]

I was thinking of using Descartes’ rule of signs, from which I find there are at most 2 positive roots and 2 negative roots of the given equation. Also, $f(\infty)>0$ and $f(0)>0$ imply that either there are no real roots or 2 real roots in $(0,\infty)$. Similar is the case for $(-\infty,0)$. If I could […]

So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$ Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work. Because there were no rational linear factors, I had […]

Problem Prove that all roots of $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real. Here $a,b \in \mathbb R$, and $a \neq 0$. Source This is one of the previous year problem of Regional Math Olympiad (India). I had a hard time solving it, so thought I’d […]

There is a general formula for solving quadratic equations, namely the Quadratic Formula. For third degree equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three equations: one for each root. Is there a general formula for solving equations of the form $ax^4+bx^3+cx^2+dx+e=0$ ? How about for higher degrees? If not, why not?

The following facts are standard: an irreducible quartic polynomial $p(x)$ can only have Galois groups $S_4, A_4, D_4, V_4, C_4$. Over a field of characteristic not equal to $2$, depending on whether or not the discriminant $\Delta$ is a square and whether or not the resolvent cubic $q(x)$ is irreducible, we can distinguish four cases: […]

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