Equation about polynomials

Question

Is it possible to find real numbers $x_0, x_1, A_0$ and $A_1$ such equation :

$$ (25a+32b+45c+80d) = A_0(ax_0^3 + bx_0^2 + cx_0 + d) + A_1(ax_1^3 + bx_1^2 + cx_1 + d) $$

is true for every real $a,b,c,d$ ?

Answer 1

So you want to solve \begin{align} 80&=A_0+A_1\\ 45&=A_0x_0+A_1x_1\\ 32&=A_0x_0^2+A_1x_1^2\\ 25&=A_0x_0^3+A_1x_1^3\\ \end{align} This is the start of the general solution formula $$ u_k=A_0x_0^k+A_1x_1^k $$ for an order 2 linear recursive sequence $u_0=80,u_1=45, u_2=32, u_3=25,…$. Solving for $p,q$ in the recursion formula $$ u_{k+2}=pu_{k+1}+qu_k $$ for the first two relations gives $$ 32=p·45+q·80\\ 25=p·32+q·45\\ $$ Since that is solvable, you can find $p$ and $q$, from that $x_0,x_1$ as solutions of the quadratic equation $$ x^2-px-q=0 $$ and from the firstmost two equations the constants $A_0,A_1$.