The following exercise can be found in the book Some exercises in pure mathematics with expository comments by authors J. D. Weston and H. J. Godwin (it is the exercise $166$ at page $43$).
Three polynomials $P$, $Q$ and $R$, of degrees $2n-1$, $2n$ and $2n+1$ respectively, are such that
$R(x)=3xQ(x)-P(x)$ for all real $x$
$P(1)=Q(1)=1$;
$P(x)$ contains only odd powers of $x$ and $Q(x)$ contains only even powers of $x$;
The equation $Q(x)=0$ has $2n$ real roots in the interval $]-1,1[$, and between each two consecutive roots lies a root of the equation $P(x)=0$.
Prove that the equation $R(x)=0$ has $2n+1$ real roots in the interval $]-1,1[$, and that between each two consecutive roots lies a root of the equation $Q(x)=0$.
What I know so far is that $R(x)$ has only odd powers of $x$ and that if you divide $R(x)$ by $3x$ you get $Q(x)$ as quotient and $P(x)$ as remaider. From here on, I have no idea on how to proceed (and don't even know if the usage of Euclid's algorithm is useful).
Edit: The exercise does not say, but I think that all roots must be distinct.
Consider the set of points A = {${-1, q_1, q_2, ... , q_{2n-1}, q_{2n}, 1}$} where $q_i$'s are the roots of Q(x).
Evaluate R(x) for each consecutive points in A. Then you will find each consecutive pairs give opposite signed results, which mean there exist a point in each consecutive pair satisfying R(p) = 0.
$R(q_i) = 3*q_i*Q(q_i) - P(q_i) = -P(q_i)$
$R(q_{i+1}) = 3*q_{i+1}*Q(q_{i+1}) - P(q_{i+1}) = - P(q_{i+1})$
Since there exists a unique root of P(x) in between each {$q_i, q_{i+1}$} , $R(q_i)$ and $R(q_{i+1})$ have opposite signs.
You should also do the same for the point pairs {$-1, q_1$} and {$q_{2n}, 1$}.
$R(-1) = -3*Q(-1) - P(-1) = -3+1 = -2$ (Q is even, P is odd)
$R(q_1) = 3*q_1*Q(q_1) - P(q_1) = -P(q_1)$
Since P has 2n-1 roots (which is odd) in ($q_1$,$q_n$) then P(1) and P(-1) have opposite signs. Thus P($q_1$) is negative, which means $R(q_1)>0$.
$R(q_{2n}) = 3*q_{2n}*Q(q_{2n}) - P(q_{2n}) = - P(q_{2n})<0 $, as there is no root of P in [$q_{2n}$,1] and they have the same sign.
$R(1) = 3*Q(1) - P(1) = 3-1 = 2$