$p$
for which the polynomial $x^4 + px^3 + px + 1$
has at least one real root.

 Find all the roots of
$x^3+x^2+x+1.$
 Sketch the graph of
$y=\frac{x^2+1}{x+1}.$
 Differentiate
$f(x)=3x^2+\frac{1}{x}.$
 Find all the roots of

Could you reformulate the question as an equation?

… and isolate
$p?$

What happens if you treat
$p$
as a function of$x?$

Why not find its range?

Have you tried sketching it?

What can you deduce from its asymptotes?

If there are any stationary points, how could you find them?

As we’re looking for roots, we’re interested in the case where the polynomial is equal to
$0.$
We rearrange to get$p=\frac{x^4+1}{x(x^2+1)}$
which we can treat as a function$p(x)$
and find its range.Notice that for
$x>0$
it is always negative and for$x<0$
always positive, with asymptotes at$x=0$
and$y=x.$
Hence we are looking for any local maxima or minima (a sketch may help to visualise it  see below).We differentiate to find
$p'(x) = \frac{x^6 + 3 x^4  3 x^2  1}{(x^3 + x)^2}$
and find the stationary points by finding the roots of$x^6 + 3 x^4  3 x^2  1.$
Since all powers are even, we can substitute$y=x^2$
and find the roots of$y^3+3y^23y1.$
Now we can notice that
$y=1$
is a root of this polynomial and factor it out to get$(y  1) (y^2 + 4 y + 1).$
The quadratic$(y^2 + 4 y + 1)$
has no real roots, so our only solutions are$x= \pm 1,$
which gives$p \leq 1$
or$p\geq1$
.