Question

A polynomial $ax^{9}+bx^{8}+1$ is divisible by $x^{2}-x-1$. Find the values of $a$ and $b$.

Spam = 10+ answers reported.
Only for quality answerers.

Answer 1

$\large\underline{\text{Step 1. Sum and product of roots}}$

Let's denote the zeros of the quadratic factor as $\alpha$ and $\beta$, then $\begin{cases} &\alpha+\beta=1\\ &\alpha\beta=-1 \end{cases}$.

$\large\underline{\text{Step 2. Zeroes of the quadratic factor}}$

If the polynomial of 9th degree is divisible by the quadratic factor

$\red{\bigstar}ax^{9}+bx^{8}+1=(x^{2}-x-1)Q(x)$

Let's substitute $x=\alpha,\beta$ since both are the zeros of the quadratic factor.

$\implies a\alpha^{9}+b\alpha^{8}=-1$

$\implies a\beta^{9}+b\beta^{8}=-1$

(by multiplying the equations by $\beta^{8}$ and $\alpha^{8}$ respectively)

$\implies a\alpha+b=-\beta^{8} \cdots\textcircled{1}$

$\implies a\beta+b=-\alpha^{8}\cdots\textcircled{2}$

$\large\underline{\text{Step 3. Finding values by identity}}$

However,

$\implies\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta=3$

$\implies\alpha^{4}+\beta^{4}=(\alpha^{2}+\beta^{2})^{2}-2(\alpha\beta)^{2}=7$

$\implies \alpha^{8}+\beta^{8}=(\alpha^{4}+\beta^{4})^{2}-2(\alpha\beta)^{4}=47$

$\large\underline{\text{Step 4. Subtraction}}$

By subtracting the two equations $\textcircled{1}$ and $\textcircled{2}$,

$\implies a(\alpha-\beta)=\alpha^{8}-\beta^{8}$

$\implies a=\dfrac{\alpha^{8}-\beta^{8}}{\alpha-\beta}$

(since $\alpha\neq\beta$)

$\implies a=(\alpha^{4}+\beta^{4})(\alpha^{2}+\beta^{2})(\alpha+\beta)$

$\implies a=7\cdot3\cdot1=21$

$\large\underline{\text{Step 5. Addition}}$

By adding the two equations $\textcircled{1}$ and $\textcircled{2}$,

$\implies a(\alpha+\beta)+2b=-\alpha^{8}-\beta^{8}$

$\implies 2b=-(\alpha^{8}+\beta^{8})-a(\alpha+\beta)$

$\implies 2b=-47-21$

$\implies b=-\dfrac{68}{2}$

$\implies b=-34$

$\large\underline{\text{Conclusion}}$

The values are $a=21,b=-34$.

$\large\underline{\text{Verification}}$

Let's try division in an actual way.

$\implies21x^{9}-34x^{8}+1=(x^{2}-x-1)(21x^{7}-13x^{6}+8x^{5}-5x^{4}+3x^{3}-2x^{2}+x-1)$

In conclusion, $21x^{9}-34x^{8}+1$ is divisible by $x^{2}-x-1$.

Answer 2

Step-by-step explanation:

given :

• A polynomial ax⁹ + bx8 + 1 is divisible by x²-x-1. Find the values of a and b.

to find :

• Find the values of a and b.

solution :

• a² + 3² = (a + b)² - 2aß = 3

• a² + ¹ = (a² + 3²)² — 2(aß)² = 7

• a + 38 = (a¹ + 3²)² — 2(aß)4 = 47

• a(a-B) = a8 - 38

• a= a- b /ab

then,

• a = = (a¹ + 34) (a² + ²)(a + B)

• a=7.3.1 = 21

• a(a +B) + 2b = - a -b

• 2b = -(a +38) - a(a + B)

• 2 b = -47 -21

• b= -34

• 21x⁹34x8 + 1 = (x² - x -

• (21x7 13x6 +8x55x4 + 3x³ - 2x² + x - 1)

• 21x⁹ – by x² - x -1.

• 34x³ + 1 is divisible

• hence , the polygon = 21x⁹ – by x² - x -1.

• 34x³ + 1 is divisible

• hope it helps you