Question

The absolute difference of the maximum and the minimum value of $x$ satisfying $x^4-16x^3+86x^2-176x+105=0$ is

Answer 1

Let $P\left(x\right)=x^4-16x^3+86x^2-176x+105$
$P\left(1\right)=1-16+86-176+105=0$
$\therefore x-1$ is a factor of $P\left(x\right)$

Using synthetic division,
$\begin{array}{cccccc}& 1& -16& 86& -176& 105\\ x=1& 0& 1& -15& 71& -105\\ & 1& -15& 71& -105& 0\end{array}$
$\therefore x^4-16x^3+86x^2-176x+105=(x-1)(x^3-15x^2+71x-105)$
Let $Q\left(x\right)=x^3-15x^2+71x-105$
By trial and error method, we can find that $x=3$ is a root of $Q\left(x\right)=0$

Now, dividing $Q\left(x\right)$ by $x-3$
$\begin{array}{ccccc}& 1& -15& 71& -105\\ x=3& 0& 3& -36& 105\\ & 1& -12& 35& 0\end{array}$
$\therefore Q\left(x\right)=(x-3)(x^2-12x+35)$
$\therefore P\left(x\right)=(x-1)(x-3)(x^2-12x+35)=(x-1)(x-3)(x-5)(x-7)$
$\therefore$ The roots of $P\left(x\right)=0$ are $1,3,5,7$

Diifference $=|7-1|=6$