A polynomial function is a function that can be written in the form:
\( P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \)
where \( n \in \mathbb{N} \cup \{0\} \) and the coefficients \( a_0, \dots, a_n \) are real numbers with \( a_n \neq 0 \).
The number 0 is called the zero polynomial.
The leading term, \( a_n x^n \), of a polynomial is the term of highest index among those terms with a non-zero coefficient.
The degree of a polynomial is the index \( n \) of the leading term.
A monic polynomial is a polynomial whose leading term has coefficient 1.
The constant term is the term of index 0. (This is the term not involving \( x \).)
A polynomial function of degree 2 is called a quadratic function. The general rule for such a function is:
\( f(x) = ax^2 + bx + c, \quad a \neq 0 \)
A polynomial function of degree 3 is called a cubic function. The general rule for such a function is:
\( f(x) = ax^3 + bx^2 + cx + d, \quad a \neq 0 \)
A polynomial function of degree 4 is called a quartic function. The general rule for such a function is:
\( f(x) = ax^4 + bx^3 + cx^2 + dx + e, \quad a \neq 0 \)
The operations of addition, subtraction, and multiplication for polynomials are naturally defined, as shown in the following examples:
Let \( P(x) = x^3 + 3x^2 + 2 \) and \( Q(x) = 2x^2 + 4 \). Then:
\( P(x) + Q(x) = (x^3 + 3x^2 + 2) + (2x^2 + 4) \)
\( = x^3 + 5x^2 + 6 \)
\( P(x) - Q(x) = (x^3 + 3x^2 + 2) - (2x^2 + 4) \)
\( = x^3 + x^2 - 2 \)
\( P(x)Q(x) = (x^3 + 3x^2 + 2)(2x^2 + 4) \)
\( = (x^3 + 3x^2 + 2) \times 2x^2 + (x^3 + 3x^2 + 2) \times 4 \)
\( = 2x^5 + 6x^4 + 4x^3 + 4x^3 + 12x^2 + 8 \)
\( = 2x^5 + 6x^4 + 4x^3 + 16x^2 + 8 \)
The sum, difference, and product of two polynomials is a polynomial.
We use the notation \(\deg(f)\) to denote the degree of a polynomial \(f\). For \(f, g \neq 0\), we have:
\(\deg(f + g) \leq \max \left( \deg(f), \deg(g) \right)\)
\(\deg(f \times g) = \deg(f) + \deg(g)\)
Two polynomials \( P \) and \( Q \) are equal only if their corresponding coefficients are equal. For two cubic polynomials:
\( P(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0 \)
\( Q(x) = b_3 x^3 + b_2 x^2 + b_1 x + b_0 \)
They are equal if and only if \( a_3 = b_3 \), \( a_2 = b_2 \), \( a_1 = b_1 \), and \( a_0 = b_0 \).
For example, if:
\( P(x) = 4x^3 + 5x^2 - x + 3 \)
\( Q(x) = b_3 x^3 + b_2 x^2 + b_1 x + b_0 \)
Then \( P(x) = Q(x) \) if and only if \( b_3 = 4 \), \( b_2 = 5 \), \( b_1 = -1 \), and \( b_0 = 3 \).
When we divide the polynomial \( P(x) \) by the polynomial \( D(x) \), we obtain two polynomials, \( Q(x) \) the quotient and \( R(x) \) the remainder, such that:
\( P(x) = D(x)Q(x) + R(x) \)
and either \( R(x) = 0 \) or \( R(x) \) has degree less than \( D(x) \).
Here \( P(x) \) is the dividend and \( D(x) \) is the divisor.
Equating polynomials is a more efficient way to factor polynomials compared to long division; moreover, the pitfalls of synthetic division could be avoided.
Suppose an \(m\)th degree polynomial, \(f(x)\), is a factor of an \(n\)th degree polynomial, \(P(x)\); then \(P(x)\) can be written as a product of \(f(x)\) and an \((n − m)\)th degree polynomial, that is,
\[ P(x) = f(x) \cdot \text{( (n − m)th degree polynomial )} \]
The coefficients of the \((n − m)\)th degree polynomial can be determined by comparing corresponding coefficients on the left and on the right side of the equation above.
Synthetic division is a simplified form of polynomial division, particularly useful for dividing a polynomial by a linear binomial of the form \(x - c\). This method is quicker and involves fewer calculations compared to long division. Below is an outline of the synthetic division process:
When dividing a polynomial by a non-linear polynomial, we use polynomial long division. The process is similar to long division with numbers, but it involves variables and their exponents. Here’s a step-by-step method for dividing by a non-linear polynomial:
Suppose that, when the polynomial \( P(x) \) is divided by \( x - \alpha \), the quotient is \( Q(x) \) and the remainder is \( R \). Then
Now, as the two expressions are equal for all values of \( x \), they are equal for \( x = \alpha \).
i.e. the remainder when \( P(x) \) is divided by \( x - \alpha \) is equal to \( P(\alpha) \). We therefore have
More generally:
When \( P(x) \) is divided by \( \beta x + \alpha \), the remainder is \( P \left( -\frac{\alpha}{\beta} \right) \)
Now, in order for \( x - \alpha \) to be a factor of the polynomial \( P(x) \), the remainder must be zero. We state this result as the factor theorem.
For a polynomial \( P(x) \):
More generally:
If \( P(x) = x^3 - a^3 \), then \( x - a \) is a factor and so by division:
\( x^3 - a^3 = (x - a)(x^2 + ax + a^2) \)
If \( a \) is replaced by \( -a \), then
\( x^3 - (-a)^3 = \)
\( x - (-a)(x^2 - ax + a^2) \)
This gives:
\( x^3 + a^3 = (x + a)(x^2 - ax + a^2) \)
Consider the cubic polynomial
\( P(x) = 2x^3 - x^2 - x - 3 \)
If the equation \( P(x) = 0 \) has a solution α that is an integer, then α divides the constant term -3. We can easily show that \( P(1) \neq 0 \), \( P(-1) \neq 0 \), \( P(3) \neq 0 \) and \( P(-3) \neq 0 \). Hence the equation \( P(x) = 0 \) has no solution that is an integer.
Does it have a rational solution, that is, a fraction for a solution?
The rational-root theorem helps us with this. It says that if α and β have highest common factor 1 (i.e. α and β are relatively prime) and βx + α is a factor of \( 2x^3 - x^2 - x - 3 \), then β divides 2 and α divides -3.
Therefore, if \( -\frac{α}{β} \) is a solution of the equation \( P(x) = 0 \) (where α and β are relatively prime), then β must divide 2 and α must divide -3. So the only value of β that needs to be considered is 2, and \( α = ±3 \) or \( α = ±1 \).
We can test these through the factor theorem. That is, check \( P(±\frac{1}{2}) \) and \( P(±\frac{3}{2}) \). We find
\( P(\frac{3}{2}) = 2(\frac{3}{2})^3 - (\frac{3}{2})^2 - \frac{3}{2} - 3 = 0 \)
We have found that \( 2x - 3 \) is a factor of \( P(x) = 2x^3 - x^2 - x - 3 \).
Dividing through, we find that
\( 2x^3 - x^2 - x - 3 = (2x - 3)(x^2 + x + 1) \)
We can show that \( x^2 + x + 1 \) has no linear factors by showing that the discriminant of this quadratic is negative.
Let \( P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \) be a polynomial of degree \( n \) with all the coefficients \( a_i \) integers. Let \( \alpha \) and \( \beta \) be integers such that the highest common factor of \( \alpha \) and \( \beta \) is 1 (i.e., \( \alpha \) and \( \beta \) are relatively prime).
If \( \beta x + \alpha \) is a factor of \( P(x) \), then \( \beta \) divides \( a_n \) and \( \alpha \) divides \( a_0 \).