Polynomial $x^2-x-1$ exactly divides Polynomial $a_1x^{17}+a_2x^{16}+1$. Calculate $a_1*a_2$

Question

Polynomial $x^2-x-1$ exactly divides Polynomial $a_1x^{17}+a_2x^{16}+1$. Calculate $a_1*a_2$

My initial thought was to substitute root from the quadratic equation into the bigger polynomial but seems like the entire degree(2) polynomial divides the bigger polynomial not its factors.

Answer 1

You're on the right track. If $(x-r)(x-s)$ divides $p(x)$, then $x-r$ and $x-s$ both divide $p(x)$ (furthermore, if $x-r$ and $x-s$ both divide $p(x)$, and $r\neq s$, then $(x-r)(x-s)$ divides $p(x)$). So, if you plug in either root of $x^2-x+1$ to $a_1x^{17}+a_2x^{16}+1$, you'll get $0$. Can you use some special properties of the roots of $x^2-x+1$ to compute $x^{16}$ and $x^{17}$ at each of the roots, and thus get equations the numbers $a_1$ and $a_2$ must satisfy?

Answer 2

I am posting the solution I worked out,

$(1+\sqrt{5})/2$ and $(1-\sqrt{5})/2$ are roots of quadratic equation.

Substituting both roots in the deg(17) polynomial we got,

$a_1*3571 +a_2*2207 +1=0$ and $-a_1*0.00028+a_2*0.00045+1=0$

solving we get $a_1 = 987$ , $a_2 = -1597$ . So $a_1*a_2 = -987*1597=-1576239$