Equality of Polynomials

 

For two polynomials to be identical (or equal), they must satisfy the following conditions:

1. The degrees of the polynomials must be equal ($\deg[P(x)] = \deg[Q(x)]$).

2. The coefficients of the corresponding terms with identical degrees must be equal.

 

Example:

 

Given the polynomials:

\[
P(x) = (a^3 + 8)x^4 – x^2 + x – 2,
\]

\[
Q(x) = (b + 1)x^3 – cx^2 + x – 2
\]

Let us find the values of $a, b,$ and $c$ assuming that $P(x) = Q(x)$.

By rewriting the polynomials to include all matching power terms with explicit coefficients (including zero coefficients), we obtain:

\[
P(x) = (a^3 + 8)x^4 + 0 \cdot x^3 – x^2 + x – 2
\]

\[
Q(x) = 0 \cdot x^4 + (b + 1)x^3 – cx^2 + x – 2
\]

Equating the corresponding coefficients of like terms yields:

\[
a^3 + 8 = 0 \Rightarrow a^3 = -8 \Rightarrow a = -2
\]

\[
0 = b + 1 \Rightarrow b = -1
\]

\[
-1 = -c \Rightarrow c = 1
\]

Thus, the parameters are uniquely determined.

 

QUESTION 8

 

\[
P(x) = 4x^4 + mx^2 + 9
\]

If the polynomial above is a perfect square trinomial, what is the negative value of $m$?

\[
\text{A)} -4 \quad
\text{B) } -6 \quad
\text{C) } -8 \quad
\text{D) } -10 \quad
\text{E) } -12
\]

 

Solution:

 

Since the leading term is a square of $2x^2$ and the constant term is a square of $3$, the expression can be modeled as a perfect square of a binomial:

\[
P(x) = 4x^4 + mx^2 + 9 = (2x^2 \pm 3)^2
\]

Expanding this expression gives:

\[
= 4x^4 \pm 12x^2 + 9
\]

By comparing the coefficients of the $x^2$ term, we find:

\[
m = \pm 12
\]

Therefore, the negative value of $m$ is $-12$.

 

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

 

QUESTION 9

 

\[
P(x) = x^4 – mx^3 + nx^2 – 6x + 1
\]

If the polynomial above is a perfect square, what is the positive value of $m$?

\[
\text{A)} 4 \quad
\text{B) } 5 \quad
\text{C) } 6 \quad
\text{D) } 7 \quad
\text{E) } 8
\]

 

Solution:

 

Since the polynomial is of degree 4, its square root must be a quadratic expression. Based on the leading term $x^4$ and the constant term $1$, we can equate it to the square of a trinomial:

\[
P(x) = x^4 – mx^3 + nx^2 – 6x + 1 = (x^2 + ax \pm 1)^2
\]

Expanding the right-hand side yields:

\[
= x^4 + 2ax^3 + (a^2 \pm 2)x^2 \pm 2ax + 1
\]

By equating the coefficients of the linear $x$ term:

\[
-6 = \pm 2a \Rightarrow a = \pm 3
\]

By equating the coefficients of the cubic $x^3$ term:

\[
-m = 2a
\]

Substituting the values of $a$ into the equation gives $m = \mp 6$. Thus, the positive value of $m$ is $6$.

 

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