Polynomial Identity Principle

Two polynomials in standard form are considered identical $$ a_nx^n + a_{n-1}x^{n-1} + \dots + a_0 = b_nx^n + b_{n-1}x^{n-1} + \dots + b_0 $$ if they have the same degree $ n $ and matching coefficients for each corresponding term: $$ a_i = b_i \quad \forall i \in \{0, 1, \dots, n\} $$

Polynomial identity is a specific type of equality.

Two polynomials are identical if they produce the same output for every input value of the variable.

In other words, they share the same algebraic structure and represent the same graph.

Note. To verify whether two polynomials are identical, express them in standard form and compare their degrees and corresponding coefficients, or check that they yield the same result for every input value. Either condition implies the other.

A practical example

Consider the polynomials:

$$ P(x) = x^3 + 2x^2 - 4x $$

$$ Q(x) = -x^3 + 3x^2 - 4x + 2x^3 - x^2 $$

Combining like terms and rewriting each polynomial in standard form, we obtain:

$$ P(x) = x^3 + 2x^2 - 4x $$

$$ Q(x) = x^3 + 2x^2 - 4x $$

Note. $$ Q(x) = -x^3 + 3x^2 - 4x + 2x^3 - x^2 $$ $$ Q(x) = (-x^3 + 2x^3) + (3x^2 - x^2) - 4x $$ $$ Q(x) = x^3 + 2x^2 - 4x $$

Example 2

Determine whether the following polynomials are identical:

$$ P(x) = 5x^2 + 4x + 3 $$

$$ Q(x) = (A + 2)x^2 + 4Bx + C $$

where $ A $, $ B $, and $ C $ are unknown coefficients.

To compare the two expressions, equate the coefficients of corresponding powers of $ x $:

$$ (A + 2)x^2 + 4Bx + C = 5x^2 + 4x + 3 $$

The coefficient of $ x^2 $ is $ A + 2 $ on the left, and 5 on the right:

A + 2 = 5

$$ \begin{cases} A + 2 = 5 \end{cases} $$

The coefficient of $ x $ is $ 4B $ on the left and 4 on the right:

4B = 4

$$ \begin{cases} A + 2 = 5 \\ 4B = 4 \end{cases} $$

The constant term is $ C $ on the left and 3 on the right:

C = 3

$$ \begin{cases} A + 2 = 5 \\ 4B = 4 \\ C = 3 \end{cases} $$

Solving this system of equations:

$$ \begin{cases} A = 5 - 2 \\ B = \frac{4}{4} \\ C = 3 \end{cases} $$

$$ \begin{cases} A = 3 \\ B = 1 \\ C = 3 \end{cases} $$

Substituting these values into $ Q(x) $, we get:

$$ Q(x) = (3 + 2)x^2 + 4(1)x + 3 = 5x^2 + 4x + 3 = P(x) $$

Hence, the two polynomials are identical.

Example 3

Find the values of $ A $, $ B $, and $ C $ that make the polynomials identical:

$$ P_1(x) = x^3 - 8x^2 + x - 3 $$

$$ P_2(x) = Cx^4 + (A + B)x^3 - Ax^2 + x - 3 $$

Setting the two polynomials equal:

$$ x^3 - 8x^2 + x - 3 = Cx^4 + (A + B)x^3 - Ax^2 + x - 3 $$

Equating coefficients of corresponding powers of $ x $:

$$ \begin{cases} C = 0 \\ A + B = 1 \\ -A = -8 \\ 1 = 1 \\ -3 = -3 \end{cases} $$

Explanation. There is no $ x^4 $ term in $ P_1 $, so its coefficient must be zero; thus, $ C = 0 $. The coefficient of $ x^3 $ is 1 in $ P_1 $ and $ A + B $ in $ P_2 $, so $ A + B = 1 $, and so forth.

We can disregard the trivial identities $ 1 = 1 $ and $ -3 = -3 $.

$$ \begin{cases} -A = -8 \\ A + B = 1 \\ C = 0 \end{cases} $$

Multiplying the first equation by -1 gives:

$$ \begin{cases} A = 8 \\ A + B = 1 \\ C = 0 \end{cases} $$

Substituting $ A = 8 $ into the second equation:

$$ \begin{cases} A = 8 \\ 8 + B = 1 \\ C = 0 \end{cases} $$

Solving for $ B $:

$$ \begin{cases} A = 8 \\ B = -7 \\ C = 0 \end{cases} $$

Substituting the values into $ P_2(x) $:

$$ P_2(x) = 0 \cdot x^4 + (8 - 7)x^3 - 8x^2 + x - 3 $$

$$ P_2(x) = x^3 - 8x^2 + x - 3 = P_1(x) $$

Thus, the polynomials are identical.

Example 4

Can the following polynomials be made identical by choosing suitable values for $ A $, $ B $, and $ C $?

$$ P_1(x) = 3x^3 + 5x^2 - 2x + 1 $$

$$ P_2(x) = Ax^4 + (A + B)x^3 - Cx^2 + x + 1 $$

Note that the coefficient of $ x $ is -2 in $ P_1 $ and +1 in $ P_2 $, but it is not parameterized in $ P_2 $.

This inconsistency means that no choice of $ A $, $ B $, or $ C $ can make the two polynomials identical.

Therefore, there is no point in solving for the parameters.

And so on.