If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ be a polynomial where a and b are real numbers and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a.
Since the function has equal positive roots, I expect $f'(\alpha) = 0$ and $f''(\alpha) > 0$ for the positive root $\alpha$ But I'm not sure how I should proceed in this question. What are the different methods of doing this problem?
The first thing to notice is that the polynomial is palindromic. Therefore if $\alpha$ is a root then so is $\frac1\alpha$.
This means that the roots must be $-x,-\frac1x, 1$ and $1$, where $x>0$ and $x\neq1$.
Considering the sum of roots, we have $$-x-\frac1x+2=-a\implies x+\frac1x=a+2$$ $$\implies a+2=(\sqrt{x}-\frac{1}{\sqrt{x}})^2+2$$
The bracketed term must be strictly positive.
Hence $$a>0$$
Note that if $r\neq 0$ is a root of $f(x)=0$, then so is $\frac{1}{r}$. This is easily noted by the symmetry of the coefficients and can algebraically shown by the fact that $$r^4f(\frac{1}{r})=f(r)$$ Hence, if the positive root is $r$ and the negative root of smaller magnitude is $s$, then our four roots are $$r,\frac{1}{r},s,\frac{1}{s}$$ However, since $r$ and $\frac{1}{r}$ are given as equal positive numbers, we have $$r=1$$ Hence, $f(x)$ is of the form $$f(x)=(x-1)^2(x-s)(x-\frac{1}{s})$$ $$f(x)=(x^2-2x+1)(x^2-(s+\frac{1}{s})x+1)$$ $$f(x)=x^4-(2+s+\frac{1}{s})x^3+(2+2s+\frac{2}{s})x^2-(2+s+\frac{1}{s})x+1$$ Hence, for all $s\neq 0$ we have that $f(x)$ fits the form $x^4+ax^3+bx^2+ax+1$.
Note that the value of $a$ is $$a=-2-s-\frac{1}{s}$$ $$a=-2+(-s)+\frac{1}{(-s)}$$ Since $s$ is given as a negative number, we have that $(-s)$ must be a positive number. Hence, by AM-GM we have $$\frac{(-s)+\frac{1}{(-s)}}{2}\geq \sqrt{(-s)\left(\frac{1}{(-s)}\right)}$$ $$(-s)+\frac{1}{(-s)}\geq 2$$ with equality when $(-s)=\frac{1}{(-s)}$. This occurs when $-s=1$. However, since this results in both negative roots being the same, equality is never attained.
Moreover, we have $$a=-2+\left((-s)+\frac{1}{(-s)}\right)\geq -2+2$$ $$a\geq 0$$ As we showed before, this equality is unattainable. The next integral value of $a$ is $\boxed{1}$. This is attainable when $$(-s)+\frac{1}{(-s)}=3$$ $$(-s)^2-3(-s)+1=0$$ $$(-s)=\frac{3\pm\sqrt{5}}{2}$$ Note that the $2$ solutions are positive, and will be the solutions to $(-s)$ and $\frac{1}{(-s)}$. Since we defined $s$ as the root with smaller magnitude, we have $s=-\frac{3-\sqrt{5}}{2}$.
Hence, the equality of $\boxed{a=1}$ is attained when $$f(x)=(x^2-2x+1)(x^2+3x+1)$$ $$f(x)=x^4+x^3-4x^2+x^3+1$$
You have
$$\begin{aligned} f(x)&=x^4+ax^3+bx^2+ax+1\\ &= x^2\left(x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}\right)\\ &=x^2\left(\left(x + \frac{1}{x}\right)^2+ a\left(x + \frac{1}{x}\right)+b-2\right) \end{aligned}$$
As $x=0$ is not a root of $f(x)$, $x$ is a root of $f(x)$ if and only if $X =x + \frac{1}{x}$ is a root of $g(X)= X^2+a X + b-2$. As all roots of $f$ are supposed to be real, so are the ones of $g$. Hence $\Delta = a^2 + 8-4b \gt0$. The roots of $g$ are
$$X_1 = \frac{-a - \sqrt{\Delta}}{2} \lt 0 \text{ and } X_2 = \frac{-a + \sqrt{\Delta}}{2} \gt 0$$
If $x_1 \lt x_2 \lt 0 \lt x_3=x_4$ are the roots of $f$, we have $X_2 = 2$ as $2$ is the only value for which the equation $x + \frac{1}{x}=X_2$ has a unique positive root. Which implies
$$a^2 + 8-4b = (4+a)^2 = 16+8a+a^2 \iff 2a+b+2=0, $$ hence $$\Delta=a^2+8-4b= a^2+8+8a+8=a^2+8a+16.$$ And as the negative roots of $f$ are distinct$$X_1 = \frac{-a - \sqrt{\Delta}}{2}\lt -2.$$ Which leads to $$a^2+8a+16 \gt (4-a)^2$$ i.e. $$16a \gt 0$$ i.e. $a \gt 0$. The minimum integral value for $a$ is therefore $1$. In that case, $$\Delta = 25, X_2=2, x_3=x_4=1$$ and $x_1,x_2$ are the values for which $$x + \frac{1}{x} = -3.$$
Two roots are positive and two negative. By Descartes rule of signs we deduce that $a\geq 0, b<0.$
Call $\alpha >0, \beta, \gamma<0$ the roots. The polynomial is palindromic, hence $f({1\over x})$ shares the roots with $f(x).$ Since the positive root is double, it is necessarilly equal to $1.$
We have $\alpha=1$ and $\gamma={1\over \beta}.$
Therefore, $$\begin{aligned}f(x)&=(x-1)^2(x-\beta)(x-{1\over \beta})\\&=(x-1)^2(x^2-(\color{blue}{\underbrace{\beta+{{1\over \beta}}}_B})+1)\\&=x^4+x^3(-2-B)+x^2(2+2B)+x(-2-B)+1\end{aligned}$$
Clearly, $a\in \mathbb{Z} \Longleftrightarrow$ $B\in \mathbb{Z}.$ The least integer $a$ could be $0.$ But if $a=0,$ then the negative root of $f(x)=0$ is also double (it is $-1$), which does not satisfy.
Let us prove that $a=1$ is convenient (it is the least theoretically convenient value). The equation $$-2-B=a=1$$ rewrites $$\beta + {1\over \beta}=-3.$$ Its (mutually reciprocal) solutions are $$\frac{-3\pm \sqrt5}{2}.$$
The given equation is $$f(x)=x^4+x^3-4x^2+x+1.$$
