Polynomials and their (real) roots 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.
 A: 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.
$R(1) = 3*Q(1) - P(1) = 3-1 = 2$
