$ax^9+bx^8+1$ is divisible by $x^2-x-1$ , where two polynomials are both functions of $x$ .

Find the value of $a$ .

*
This problem is a part of
<Grade 10 CSAT Mock test> series
.
*

The answer is 21.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Let $p\text{, }q$ be roots of $x^2-x-1=0$ . Then, $p+q=1$ and $pq=-1$ .

Therefore, $p^2+q^2=3$ and $p^4+q^4=7$ .

Since $ax^9+bx^8+1=(x^2-x-1)Q(x)$ ,

$ap^9+bp^8=-1 \text{ }\cdots\text{ }A\\ aq^9+bq^8=-1 \text{ }\cdots\text{ }B$

Multiply $q^8$ to both sides of $A$ , and $p^8$ to both sides of $B$ .

$ap(pq)^8+b(pq)^8=-q^8 \\ aq(pq)^8+b(pq)^8=-p^8$

Since $pq=-1$ ,

$ap+b=-q^8 \\ aq+b=-p^8$

Subtract the lower equation from the upper equation and you get

$\begin{aligned} a(p-q) & =p^8-q^8 \\ a & =\frac{p^8-q^8}{p-q} \\ & =\frac{(p^4+q^4)(p^2+q^2)(p+q)(p-q)}{(p-q)} \\ & =7\times3\times1 \\ & =\boxed{21} \\ \end{aligned}$