$p$ 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$y^3+3y^2-3y-1.$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$p\geq1$.