Question

The ordered pair (a, b) such that $x^2 - x - 1$ divides $ax^9 + bx^8 + 1$ is

• A. (-21,34)
• B. (21,-34)
• C. (47,-58)
• D. (58,-47)

Answer 1

Answer: (B)

(21,-34)

Answer 2

If a polynomial $P(x)$ divides another polynomial $Q(x)$, then all the roots of $P(x)$ must also be roots of $Q(x)$.

First, find the roots of the divisor polynomial $x^2 - x - 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Let $\alpha = \frac{1 + \sqrt{5}}{2}$ be one of the roots. Since $\alpha$ is a root of $x^2 - x - 1 = 0$, it satisfies the equation $\alpha^2 - \alpha - 1 = 0$, which implies $\alpha^2 = \alpha + 1$.

Next, we need to evaluate powers of $\alpha$. The powers of $\alpha$ follow a pattern related to the Fibonacci sequence. Let $F_n$ be the Fibonacci sequence defined by $F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ for $n \ge 2$. The sequence starts: $0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \dots$ For a root $\alpha$ of $x^2 - x - 1 = 0$, it is a known property that $\alpha^n = F_n \alpha + F_{n-1}$. Let's list the required Fibonacci numbers: $F_0 = 0$ $F_1 = 1$ $F_2 = 1$ $F_3 = 2$ $F_4 = 3$ $F_5 = 5$ $F_6 = 8$ $F_7 = 13$ $F_8 = 21$ $F_9 = 34$

Using the formula: $\alpha^8 = F_8 \alpha + F_7 = 21\alpha + 13$. $\alpha^9 = F_9 \alpha + F_8 = 34\alpha + 21$.

Since $x^2 - x - 1$ divides $ax^9 + bx^8 + 1$, the root $\alpha$ must satisfy $a\alpha^9 + b\alpha^8 + 1 = 0$. Substitute the expressions for $\alpha^8$ and $\alpha^9$: $a(34\alpha + 21) + b(21\alpha + 13) + 1 = 0$ Expand and group terms by $\alpha$: $(34a + 21b)\alpha + (21a + 13b + 1) = 0$

Since $\alpha = \frac{1 + \sqrt{5}}{2}$ is an irrational number, and $a, b$ are rational coefficients, for this equation to hold, both the coefficient of $\alpha$ and the constant term must be zero. This gives us a system of two linear equations:

1. $34a + 21b = 0$
2. $21a + 13b + 1 = 0$

From equation (1), we can express $b$ in terms of $a$: $21b = -34a \implies b = -\frac{34}{21}a$.

Substitute this expression for $b$ into equation (2): $21a + 13\left(-\frac{34}{21}a\right) + 1 = 0$ $21a - \frac{442}{21}a + 1 = 0$ Multiply the entire equation by 21 to clear the denominator: $21(21a) - 442a + 21(1) = 0$ $441a - 442a + 21 = 0$ $-a + 21 = 0$ $a = 21$

Now substitute the value of $a$ back into the expression for $b$: $b = -\frac{34}{21}(21)$ $b = -34$

Thus, the ordered pair $(a, b)$ is $(21, -34)$.