Question

If $P$ is the positive integer for which $1+x^4+x^8+\dots \dots x^{4(p-1)}$ is divisible $(1+x+\dots ..+x^{p-1})$ and both the roots of $f(x)=x^2-(p+6)x-2q=0$ are prime numbers, then the value of $p+q+2$ is (1) $-1$ (2) 2 (3) $-2$ (4) 0

Short answer

The value is 11.

Step-by-step solution

- Analyzing the divisibility condition

Observe the polynomial expression: $1+x^4+x^8+\text{rdots}+x^{4(p-1)}$. We need this to be divisible by $(1+x+x^2+\text{rdots}+x^{p-1})$. This implies that $P$ should be such that every power of $x$ in the numerator can be written as some power found in the denominator.

- Applying geometric series properties

The expression $1+x^4+x^8+\text{rdots}+x^{4(p-1)}$ is a geometric series with the first term 1 and common ratio $x^4$. The sum of this series can be written as $\frac{1-x^{4p}}{1-x^4}$. Similarly, $1+x+x^2+\text{rdots}+x^{p-1}$ is $\frac{1-x^p}{1-x}$.

- Relating the series for divisibility

For $\frac{1-x^{4p}}{1-x^4}$ to be divisible by $\frac{1-x^p}{1-x}$, we need $4p=mp$ for some integer $m$. This implies the smallest positive integer value for $m$ is 4. Hence, $p$ must be 4.

- Solving the quadratic equation

Given the function $f(x)=x^2-(p+6)x-2q$, and knowing both roots are prime numbers, we substitute $p=4$, leading to $f(x)=x^2-10x-2q=0$. Consider prime roots such as 2 and 5. We form the equation product: $(x-2)(x-5)=x^2-7x+10$. Hence, $10=2q$.

- Solving for $q$

Comparing coefficients, we find $2q=10$, so $q=5$.

- Calculating the final expression

Finally, the question asks for the value of $p+q+2$. Given $p=4$ and $q=5$, we calculate: $p+q+2=4+5+2=11$.

Key concepts

Divisibility in Polynomials

Divisibility in polynomials is a crucial concept in understanding how one polynomial can be divided by another without leaving a remainder. In our exercise, we examine the polynomial expression:

• $1+x^4+x^8+\dots +x^{4(p-1)}$

Here, we require it to be divisible by another polynomial:

• $(1+x+x^2+\dots +x^{p-1})$

To ensure that the polynomial is divisible, the terms in the numerator must align with the terms in the denominator. We explored this condition and figured out that $4p$ should align with $mp$, where $m$ is an integer. This leads us to the conclusion that $p$ should be 4.

Geometric Series

A geometric series is a sequence of terms where each term is a fixed multiple of the previous term. In our exercise, the series is:

• $1+x^4+x^8+\dots +x^{4(p-1)}$

• This can be summed up using the geometric series formula: $\sum _{k=0}^{n}a{r}^k=\frac{a(1-r^{n+1})}{1-r}$

Here, $a=1$ and the common ratio $r=x^4$. Therefore, the series sums to $\frac{1-x^{4p}}{1-x^4}$. Similarly, the series $1+x+x^2+\dots +x^{p-1}$ sums to $\frac{1-x^p}{1-x}$. For the polynomial divisibility condition, we noted that $\frac{1-x^{4p}}{1-x^4}$ must align with $\frac{1-x^p}{1-x}$, solidifying $p$ as 4.

Solving Quadratic Equations with Prime Roots

Finally, solving quadratic equations with prime roots requires understanding the fundamentals of quadratic equations. The general form of the quadratic equation in this exercise is:

• $f(x)=x^2-(p+6)x-2q$

We were given that both roots must be prime numbers. After setting $p=4$, we get:

• $f(x)=x^2-(4+6)x-2q=x^2-10x-2q$

We then consider potential prime roots like $2$ and $5$. Factoring them into the equation, we find:

• $(x-2)(x-5)=x^2-7x+10$

Comparing with the standard form, we get $10=2q$, giving us $q=5$. The final value of $p+q+2$ results in $4+5+2=11$. This shows the solution and relationship between quadratic equations and prime number roots.