#20 of June 2016 Grade 10 CSAT (Korean SAT) mock test

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

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

When $x^{2} - x-1=0$ , $\Rightarrow x^2=x+1$ $x^4=(x+1)^2=x^2+2x+1=(x+1) +2x+1=3x+2$ $x^8=(3x+2)^2=9x^2+12x+4=9(x+1)+12x+4=21x+13$ $x^9=x(21x+13)=21x^2+13=21(x+1)+13=34x+21$

Since $ax^{9}+bx^{8}+1$ is divisible by $x^2-x-1$ , we deduce that these two polynomials have common roots.

Hence, by substitution

$ax^9+bx^8+1=0$ can be rewritten as $a(34x+21)+b(21x+13)+1=0$ Equating the coefficients of each term, we get $\begin{cases} 34a+21b=0\\21a+13b+1=0\end{cases}$

Solving the system of simultaneous equations, we get $a=21,b=-34$ .

Here's a different approach

Just use polynomial division. And equate the remainder to zero,because it completely divides. You get 2 simultaneous linear equations. Solve them to get A.