Answer to Question #262420 in Math for Shah

Question

A polynomials is defined by P(x) = mx³ + mx² + mx + n where m and n are non-zero real constants. Given that x = -n is a root of P(x). Determine the range of the possible values of m.

Answer 1

$$P(-n)=-mn^3+mn^2-mn+n=0$$

$n\not=0$

$$-mn^2+mn-m+1=0$$

$$m=\dfrac{1}{n^2-n+1}$$

Let $m=f(n)=\dfrac{1}{n^2-n+1}, mn\not=0$

$$n^2-n+1=n^2-n+\dfrac{1}{4}+\dfrac{3}{4}$$

$$=(n-\dfrac{1}{2})^2+\dfrac{3}{4}>0,n\in \R$$

$$f'(n)=-\dfrac{2n-1}{(n^2-n+1)^2}$$

Find the critical number(s)

$$f'(n)=0=>-\dfrac{2n-1}{(n^2-n+1)^2}=0=>n=\dfrac{1}{2}$$

If $n\in(-\infin, 0)\cup(0, \dfrac{1}{2}),$ then $f'(n)>0, f(n)$ increases.

If $n\in(\dfrac{1}{2},\infin),$ then $f'(n)<0, f(n)$ decreases.

The function $f(n)$ has a local maximum at $n=\dfrac{1}{2}.$

$$f(\dfrac{1}{2})=\dfrac{1}{(\dfrac{1}{2})^2-\dfrac{1}{2}+1}=\dfrac{4}{3}$$

$$\lim\limits_{n\to0^-}f(n)=\lim\limits_{n\to0^+}f(n)=1$$

$$\lim\limits_{n\to-\infin}f(n)=\lim\limits_{n\to\infin}f(n)=0$$

Then $m\in(0, \dfrac{4}{3}]$