If xn - pyn + qzn is divisible by x2 + abyz - bzx - axy, then what is $\rm \frac{p}{a^n}-\frac{q}{b^n}$ equal to ?

Question

Prepp Team · Accepted Answer

Understanding the Polynomial Divisibility Problem

The question asks for the value of a specific expression involving $p$, $q$, $a$, and $b$, given that the polynomial $x^n - py^n + qz^n$ is divisible by another polynomial $x^2 + abyz - bzx - axy$.

When one polynomial is divisible by another, it means that the divisor is a factor of the dividend. If a polynomial $P(x)$ is divisible by $(x-\alpha)$, then $P(\alpha)=0$. This concept extends to multivariate polynomials and multiple factors.

Factoring the Divisor Polynomial

The divisor polynomial is given as $x^2 + abyz - bzx - axy$. Let's try to factor this expression.

We can rearrange the terms and group them:

$x^2 - axy - bzx + abyz$

Group the first two terms and the last two terms:

$(x^2 - axy) - (bzx - abyz)$

Factor out common terms from each group:

$x(x - ay) - bz(x - ay)$

Now, we can see that $(x - ay)$ is a common factor:

$(x - ay)(x - bz)$

So, the divisor polynomial is $(x - ay)(x - bz)$.

Applying the Divisibility Condition

If the polynomial $x^n - py^n + qz^n$ is divisible by $(x - ay)(x - bz)$, it must be divisible by each of the factors, $(x - ay)$ and $(x - bz)$, assuming these factors are relatively prime (which they are, unless $ay=bz$).

According to the factor theorem principle for multivariate polynomials, if a polynomial $P(x,y,z)$ is divisible by $(x - k)$, then $P(k,y,z)$ must be the zero polynomial. Extending this, if $P(x,y,z)$ is divisible by $(x - ay)$, then $P(ay,y,z)$ must be the zero polynomial (in terms of $y$ and $z$). Similarly, if it's divisible by $(x - bz)$, then $P(bz,y,z)$ must be the zero polynomial.

Let $P(x,y,z) = x^n - py^n + qz^n$.

Condition 1: Divisibility by $(x - ay)$

Substitute $x = ay$ into the polynomial $P(x,y,z)$:

$P(ay, y, z) = (ay)^n - py^n + qz^n$

$P(ay, y, z) = a^n y^n - py^n + qz^n$

For divisibility by $(x - ay)$, this expression must be zero:

$a^n y^n - py^n + qz^n = 0$

We can group the terms with $y^n$:

$(a^n - p)y^n + qz^n = 0$ (Equation 1)

Condition 2: Divisibility by $(x - bz)$

Substitute $x = bz$ into the polynomial $P(x,y,z)$:

$P(bz, y, z) = (bz)^n - py^n + qz^n$

$P(bz, y, z) = b^n z^n - py^n + qz^n$

For divisibility by $(x - bz)$, this expression must be zero:

$b^n z^n - py^n + qz^n = 0$

Rearrange the terms:

$-py^n + (b^n + q)z^n = 0$ (Equation 2)

Solving the System of Equations

We have two linear equations involving $y^n$ and $z^n$:

1) $(a^n - p)y^n + qz^n = 0$

2) $-py^n + (b^n + q)z^n = 0$

For these two homogeneous linear equations in $y^n$ and $z^n$ to have a non-trivial solution (i.e., solutions other than $y^n = 0$ and $z^n = 0$, which implies $y=0$ and $z=0$), the determinant of the coefficient matrix must be zero.

The coefficient matrix is:

\[\begin{pmatrix} a^n - p & q \\ -p & b^n + q \end{pmatrix}\]

The determinant of this matrix is calculated as:

\[\text{Determinant} = (a^n - p)(b^n + q) - (q)(-p)\]

Set the determinant equal to zero:

\[(a^n - p)(b^n + q) + pq = 0\]

Expand the product:

\[a^n b^n + a^n q - p b^n - pq + pq = 0\]

The $-pq$ and $+pq$ terms cancel out:

\[a^n b^n + a^n q - p b^n = 0\]

We want to find the value of $\frac{p}{a^n} - \frac{q}{b^n}$. Let's rearrange the equation to isolate terms involving $p$ and $q$.

Move the terms with $p$ and $q$ to one side:

\[p b^n - a^n q = a^n b^n\]

Assuming $a \ne 0$ and $b \ne 0$, we can divide the entire equation by $a^n b^n$:

\[\frac{p b^n}{a^n b^n} - \frac{a^n q}{a^n b^n} = \frac{a^n b^n}{a^n b^n}\]

Simplify the terms:

\[\frac{p}{a^n} - \frac{q}{b^n} = 1\]

Thus, the value of $\frac{p}{a^n} - \frac{q}{b^n}$ is 1.

Conclusion

Based on the divisibility condition and solving the resulting algebraic equations, we found that the value of $\frac{p}{a^n} - \frac{q}{b^n}$ is 1.

Revision Table: Key Concepts

Here is a summary of the key mathematical concepts used in solving this polynomial divisibility problem:

Polynomial Factoring: Breaking down a polynomial into a product of simpler polynomials (e.g., factoring $x^2 - axy - bzx + abyz$).
Polynomial Divisibility: The property that one polynomial can be divided by another with no remainder. If $P(x)$ is divisible by $D(x)$, then $P(x) = Q(x)D(x)$ for some polynomial $Q(x)$.
Factor Theorem: For a polynomial $P(x)$, $(x - c)$ is a factor if and only if $P(c) = 0$. This extends to multivariate polynomials; if $P(x,y,z)$ is divisible by $(x-ay)$, then $P(ay,y,z)$ is the zero polynomial in $y$ and $z$.
System of Linear Equations: A set of two or more linear equations involving the same variables.
Determinant of a Matrix: A scalar value calculated from the elements of a square matrix. For a system of homogeneous linear equations to have non-trivial solutions, the determinant of the coefficient matrix must be zero.

Additional Information: Polynomial Properties

Understanding polynomial properties is crucial for solving such problems. Here are a few related points:

Homogeneous Polynomials: A polynomial where all terms have the same total degree. In this problem, $x^n - py^n + qz^n$ is homogeneous of degree $n$, and the divisor $(x-ay)(x-bz)$ is homogeneous of degree 2. If a homogeneous polynomial is divisible by another homogeneous polynomial, the quotient is also homogeneous.
Roots of a Polynomial: Values of the variables that make the polynomial equal to zero. The factor theorem links factors to roots. The roots of $(x-ay)(x-bz)=0$ with respect to $x$ (for fixed $y,z$) are $x=ay$ and $x=bz$. The divisibility implies these are also roots of $x^n - py^n + qz^n$ when considered as a polynomial in $x$.
Linear Dependence: Two equations $A_1 X + B_1 Y = 0$ and $A_2 X + B_2 Y = 0$ are linearly dependent if one is a constant multiple of the other, or if the determinant $A_1 B_2 - A_2 B_1 = 0$. If they are linearly dependent, they have infinitely many solutions or only the trivial solution, but the determinant condition specifically allows for non-trivial solutions if they exist. In our case, the conditions derived must hold for any valid values of $y, z$, which implies a dependency between the resulting equations in $y^n$ and $z^n$.

Prepp Team · Answer

-1

Prepp Team · Answer

0

Prepp Team · Answer

2

Understanding the Polynomial Divisibility Problem

Factoring the Divisor Polynomial

Applying the Divisibility Condition

Condition 1: Divisibility by \((x - ay)\)

Condition 2: Divisibility by \((x - bz)\)

Solving the System of Equations

Conclusion

Revision Table: Key Concepts

Additional Information: Polynomial Properties

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