Multivariate Polynomials

 

Let $x, y,$ and $z$ be variable elements. Expressions in the form of

\[
P(x, y) = x^3 y^2 + 2xy^3 – xy + x + y + 1
\]

are defined as bivariate polynomials (polynomials in two variables), while expressions in the form of

\[
P(x, y, z) = x^2 yz^2 – xy^2 + xz + x – z + 3
\]

are defined as trivariate polynomials (polynomials in three variables).

In a multivariate polynomial, the degree of a specific term is the sum of the exponents of all the variables contained within that term. The total degree of the polynomial corresponds to the highest degree among all of its constituent terms.

To find the sum of the coefficients of a multivariate polynomial, evaluate the polynomial by substituting $1$ for all variables. To find the constant term, evaluate the polynomial by substituting $0$ (zero) for all variables.

 

Examples:

 

$\bullet \quad$ For the polynomial $P(x, y) = x^4 y^2 + x^5 – y^4 + xy + 3$, the degree is determined by its leading term $x^4 y^2$:

\[ 4 + 2 = 6 \] \[\deg[P(x, y)] = 6. \]

 

$\bullet \quad$ For the polynomial $P(x, y, z) = x^3 y^2 z + xyz – xy + x + z + 5$, the degree is determined by its leading term $x^3 y^2 z$:

\[ 3 + 2 + 1 = 6 \] \[\deg[P(x, y, z)] = 6. \]

 

$\bullet \quad$ For the polynomial $P(x, y, z) = x^2 yz – 2x^2 + 3yz + z + 3$, the sum of the coefficients is found via:
\[
P(1,1,1) = 1 – 2 + 3 + 1 + 3 = 6
\]

$\bullet \quad$ The constant term is found via:
\[
P(0,0,0) = 3.
\]

 

QUESTION 7

 

\[
P(x, y) = \frac{xy^3 + y^3 + xy + y^2 + y + 1}{y^2 + 1}
\]

Given that the expression above represents a bivariate polynomial, what is the sum of the coefficients of the polynomial $P(x – 1, y + 1)$?

\[
\text{A)} 6 \quad
\text{B) } 5 \quad
\text{C) } 4 \quad
\text{D) } 3 \quad
\text{E) } 2
\]

 

Solution:

 

The sum of the coefficients of the polynomial $P(x – 1, y + 1)$ is found by substituting $x = 1$ and $y = 1$:

\[ P(1 – 1, \;\; 1 + 1) = P(0, 2). \]

Evaluating the given bivariate polynomial expression

\[
P(x, y) = \frac{xy^3 + y^3 + xy + y^2 + y + 1}{y^2 + 1}
\]

at $x = 0$ and $y = 2$ directly yields $P(0,2) = 3$.

 

\(\textbf{Correct Answer: D} \)