AOS1 Topic 4: Polynomial Functions

Basic definitions

Polynomial Function

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 \( a_n \neq 0 \).

Zero Polynomial

The number 0 is called the zero polynomial.

Leading Term

The leading term, \( a_n x^n \), of a polynomial is the term of highest index among those terms with a non-zero coefficient.

Degree of a Polynomial

The degree of a polynomial is the index \( n \) of the leading term.

Monic Polynomial

A monic polynomial is a polynomial whose leading term has coefficient 1.

Constant Term

The constant term is the term of index 0. (This is the term not involving \( x \).)

Polynomial Functions

Quadratic Function

A polynomial function of degree 2 is called a quadratic function. The general rule for such a function is:

\( f(x) = ax^2 + bx + c, \quad a \neq 0 \)

Cubic Function

A polynomial function of degree 3 is called a cubic function. The general rule for such a function is:

\( f(x) = ax^3 + bx^2 + cx + d, \quad a \neq 0 \)

Quartic Function

A polynomial function of degree 4 is called a quartic function. The general rule for such a function is:

\( f(x) = ax^4 + bx^3 + cx^2 + dx + e, \quad a \neq 0 \)

Arithmetic of Polynomials

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.

Note:

We use the notation \(\deg(f)\) to denote the degree of a polynomial \(f\). For \(f, g \neq 0\), we have:

\(\deg(f + g) \leq \max \left( \deg(f), \deg(g) \right)\)

\(\deg(f \times g) = \deg(f) + \deg(g)\)

Equating Coefficients

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 \( b_0 = 3 \).

Division and Factorisation of Polynomials

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 Coefficients Method for Polynomial Division

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.

Division of Polynomials by Synthetic Division

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:

  1. Setup: Write down the coefficients of the polynomial you want to divide. If any terms are missing, use a zero as the coefficient for that term.
  2. Division: Place the value of \(c\) (from \(x - c\)) to the left, outside the division symbol. Draw a horizontal line, and below it, write the coefficients from step 1.
  3. Bring down the leading coefficient: Bring down the leading coefficient (the first number) to the bottom row.
  4. Multiply and add: Multiply the value you just brought down by \(c\) and write the result under the next coefficient. Add the numbers in this column and write the result below the line.
  5. Repeat: Repeat the multiply and add process for each coefficient. The final row of numbers represents the coefficients of the quotient polynomial, with the last number being the remainder.

Dividing by a Non-linear Polynomial

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:

  1. Set up the division: Write the dividend (the polynomial to be divided) and the divisor (the non-linear polynomial) in standard form, with terms ordered by decreasing exponents.
  2. Divide the leading terms: Divide the leading term of the dividend by the leading term of the divisor. Write the result above the division line; this is the first term of the quotient.
  3. Multiply and subtract: Multiply the entire divisor by the term obtained in the previous step. Subtract this product from the dividend to form a new polynomial.
  4. Repeat: Repeat the divide, multiply, and subtract steps with the new polynomial formed until the degree of the remainder is less than the degree of the divisor.

The Remainder Theorem

Suppose that, when the polynomial \( P(x) \) is divided by \( x - \alpha \), the quotient is \( Q(x) \) and the remainder is \( R \). Then

\( P(x) = (x - \alpha)Q(x) + R \)

Now, as the two expressions are equal for all values of \( x \), they are equal for \( x = \alpha \).

\( \therefore P(\alpha) = (\alpha - \alpha)Q(\alpha) + R \therefore R = P(\alpha) \)

i.e. the remainder when \( P(x) \) is divided by \( x - \alpha \) is equal to \( P(\alpha) \). We therefore have

\( P(x) = (x - \alpha)Q(x) + P(\alpha) \)

More generally:

Remainder Theorem

When \( P(x) \) is divided by \( \beta x + \alpha \), the remainder is \( P \left( -\frac{\alpha}{\beta} \right) \)

The Factor Theorem

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.

Factor Theorem

For a polynomial \( P(x) \):

  • If \( P(\alpha) = 0 \), then \( x - \alpha \) is a factor of \( P(x) \).
  • Conversely, if \( x - \alpha \) is a factor of \( P(x) \), then \( P(\alpha) = 0 \).

More generally:

  • If \( \beta x + \alpha \) is a factor of \( P(x) \), then \( P \left( -\frac{\alpha}{\beta} \right) = 0 \).
  • Conversely, if \( P \left( -\frac{\alpha}{\beta} \right) = 0 \), then \( \beta x + \alpha \) is a factor of \( P(x) \).

Sums and Differences of Cubes

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:

\( x^3 + a^3 = (x + a)(x^2 - ax + a^2) \)

The Rational-Root Theorem

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.

Rational-Root Theorem

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 \( a_0 \).

Example 1

Polynomial Evaluation

Let \( P(x) = x^4 - 3x^3 - 2 \). Find:

a) \( P(1) \)

b) \( P(-1) \)

Solution:

\( P(1) = (1)^4 - 3(1)^3 - 2 \)
\( = 1 - 3 - 2 \)
\( = -4 \)

Solution:

\( P(-1) = (-1)^4 - 3(-1)^3 - 2 \)
\( = 1 + 3 - 2 \)
\( = 2 \)

Example 2

Polynomial Evaluation

Let \( Q(x) = 2x^6 - x^3 + ax^2 + bx + 20 \). If \( Q(-1) = Q(2) = 0 \), find the values of \( a \) and \( b \).

Solution:

\( Q(-1) = 2(-1)^6 - (-1)^3 + a(-1)^2 - b + 20 \)
\( = 2 + 1 + a - b + 20 \)
\( = 23 + a - b \)

\( Q(2) = 2(2)^6 - (2)^3 + a(2)^2 + 2b + 20 \)
\( = 128 - 8 + 4a + 2b + 20 \)
\( = 140 + 4a + 2b \)

First find \( Q(-1) \) and \( Q(2) \) in terms of \( a \) and \( b \).


Since \( Q(-1) = Q(2) = 0 \), this gives:
\( 23 + a - b = 0 \)      (1)
\( 140 + 4a + 2b = 0 \)      (2)

Divide (2) by 2:
\( 70 + 2a + b = 0 \)      (3)

Add (1) and (3):
\( 93 + 3a = 0 \)
\( \therefore a = -31 \)

Substitute in (1) to obtain \( b = -8 \).
Form simultaneous equations in \( a \) and \( b \) by putting \( Q(-1) = 0 \) and \( Q(2) = 0 \).
Example 3

Polynomial Operations

Let \( P(x) = x^3 - 6x + 3 \) and \( Q(x) = x^2 - 3x + 1 \). Find:

a.\( P(x) - Q(x) \)

b. \( P(x)Q(x) \)

Solution:

\( P(x) - Q(x) = (x^3 - 6x + 3) - (x^2 - 3x + 1) \)
\( P(x) - Q(x) = x^3 - 6x + 3 - x^2 + 3x - 1 \)
\( P(x) - Q(x) = x^3 - x^2 - 3x + 2 \)

\( P(x)Q(x) = (x^3 - 6x + 3)(x^2 - 3x + 1) \)
Using the distributive property:
\( P(x)Q(x) = x^3(x^2 - 3x + 1) - 6x(x^2 - 3x + 1) + 3(x^2 - 3x + 1) \)
\( P(x)Q(x) = (x^5 - 3x^4 + x^3) - (6x^3 - 18x^2 + 6x) + (3x^2 - 9x + 3) \)
Combine like terms:
\( P(x)Q(x) = x^5 - 3x^4 + x^3 - 6x^3 + 18x^2 - 6x + 3x^2 - 9x + 3 \)
\( P(x)Q(x) = x^5 - 3x^4 - 5x^3 + 21x^2 - 15x + 3 \)

Example 4

Equating coefficients:

1. If \( x^3 + 3x^2 + 3x + 8 = a(x + 1)^3 + b \) for all \( x \in \mathbb{R} \), find the values of \( a \) and \( b \).

>2.Show that \( x^3 + 6x^2 + 6x + 8 \) cannot be written in the form \( a(x + c)^3 + b \) for real numbers \( a, b \) and \( c \).

Solution:

Expand the right-hand side of the equation:

\( a(x + 1)^3 + b = a(x^3 + 3x^2 + 3x + 1) + b \)

\( = ax^3 + 3ax^2 + 3ax + a + b \)

If \( x^3 + 3x^2 + 3x + 8 = ax^3 + 3ax^2 + 3ax + a + b \) for all \( x \in \mathbb{R} \), then by equating coefficients:

  • Coefficient of \( x^3 \): \( 1 = a \)
  • Coefficient of \( x^2 \): \( 3 = 3a \)
  • Coefficient of \( x \): \( 3 = 3a \)
  • Constant term: \( 8 = a + b \)

Hence, \( a = 1 \) and \( b = 7 \).

Solution 2:

Expand the proposed form:

\( a(x + c)^3 + b = a(x^3 + 3cx^2 + 3c^2x + c^3) + b \)

\( = ax^3 + 3cax^2 + 3c^2ax + c^3a + b \)

Suppose \( x^3 + 6x^2 + 6x + 8 = ax^3 + 3cax^2 + 3c^2ax + c^3a + b \) for all \( x \in \mathbb{R} \). Then:

  • Coefficient of \( x^3 \): \( 1 = a \) (1)
  • Coefficient of \( x^2 \): \( 6 = 3ca \) (2)
  • Coefficient of \( x \): \( 6 = 3c^2a \) (3)
  • Constant term: \( 8 = c^3a + b \) (4)

From (1), we have \( a = 1 \). So from (2), we have \( c = 2 \).

But substituting \( a = 1 \) and \( c = 2 \) into (3) gives \( 6 = 12 \), which is a contradiction.

Example 5

Polynomial Division

Divide \(x^3 + x^2 - 14x - 24\) by \(x + 2\).

Solution:

\[ \begin{array}{r|rrr} x + 2 & x^2 - x - 12 \\ \hline x^3 + x^2 - 14x - 24 & x^3 + 2x^2 \\ & -x^2 - 14x - 24 \\ & -x^2 - 2x \\ & -12x - 24 \\ & -12x - 24 \\ & 0 \\ \end{array} \]


1. Divide \(x\) from \(x + 2\) into the leading term \(x^3\) to get \(x^2\).

2. Multiply \(x^2\) by \(x + 2\) to give \(x^3 + 2x^2\).

3. Subtract from \(x^3 + x^2 - 14x - 24\), leaving \(-x^2 - 14x - 24\).

4. Now divide \(x\) from \(x + 2\) into \(-x^2\) to get \(-x\).

5. Multiply \(-x\) by \(x + 2\) to give \(-x^2 - 2x\).

6. Subtract from \(-x^2 - 14x - 24\), leaving \(-12x - 24\).

7. Divide \(x\) into \(-12x\) to get \(-12\).

8. Multiply \(-12\) by \(x + 2\) to give \(-12x - 24\).

9. Subtract from \(-12x - 24\), leaving a remainder of 0.

In this example, we see that \(x + 2\) is a factor of \(x^3 + x^2 - 14x - 24\), as the remainder is zero.

Thus, \((x^3 + x^2 - 14x - 24) \div (x + 2) = x^2 - x - 12\) with zero remainder.

Example 6

Finding the Factors of a Polynomial

Show that \( x + 1 \) is a factor of \( x^3 − 4x^2 + x + 6 \) and hence find the other linear factors.

Solution:

Let \( P(x) = x^3 − 4x^2 + x + 6 \).

Evaluate \( P(-1) \):

\[ P(-1) = (-1)^3 − 4(-1)^2 + (-1) + 6 = 0 \]

Thus, \( x + 1 \) is a factor (by the factor theorem).

Divide by \( x + 1 \) to find the other factor:

\[ \begin{array}{r|rrrr} & x^2 & -5x & +6 \\ \hline x+1 & x^3 & -4x^2 & +x & +6 \\ & x^3 & +x^2 \\ \hline & -5x^2 & +x & +6 \\ & -5x^2 & -5x \\ \hline & 6x & +6 \\ & 6x & +6 \\ \hline & 0 \\ \end{array} \]

Therefore:

\[ x^3 − 4x^2 + x + 6 = (x + 1)(x^2 − 5x + 6) = (x + 1)(x − 3)(x − 2) \]

The linear factors of \( x^3 − 4x^2 + x + 6 \) are \( (x + 1) \), \( (x − 3) \), and \( (x − 2) \).

Explanation:

We can use the factor theorem to find one factor, and then divide to find the other two linear factors.

Here is an alternative method:

Once we have found that \( x + 1 \) is a factor, we know that we can write:

\[ x^3 - 4x^2 + x + 6 = (x + 1)(x^2 + bx + c) \]

By equating constant terms, we have:

\[ 6 = 1 \times c \implies c = 6 \]

By equating coefficients of \( x^2 \), we have:

\[ -4 = 1 + b \implies b = -5 \]

Therefore:

\[ x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6) \]

Example 7

Factorize this \(P(x) = 3x^3 + 8x^2 + 2x - 5\) Using the Rational-Root Theorem

Solution:

First, we test the possible rational roots using the Rational-Root Theorem:

\( P(1) = 3(1)^3 + 8(1)^2 + 2(1) - 5 = 8 \neq 0 \)
\( P(-1) = 3(-1)^3 + 8(-1)^2 + 2(-1) - 5 = -2 \neq 0 \)
\( P(5) = 3(5)^3 + 8(5)^2 + 2(5) - 5 = 580 \neq 0 \)
\( P(-5) = 3(-5)^3 + 8(-5)^2 + 2(-5) - 5 = -190 \neq 0 \)
\( P\left(-\frac{5}{3}\right) = 3\left(-\frac{5}{3}\right)^3 + 8\left(-\frac{5}{3}\right)^2 + 2\left(-\frac{5}{3}\right) - 5 = 0 \)

Therefore, \(3x + 5\) is a factor.

Dividing \(P(x)\) by \(3x + 5\), we get:

\( 3x^3 + 8x^2 + 2x - 5 = (3x + 5)(x^2 + x - 1) \)

Next, we complete the square for \(x^2 + x - 1\) to factorise it:

\( x^2 + x - 1 = x^2 + x + \frac{1}{4} - \frac{1}{4} - 1 \)
\( = \left(x + \frac{1}{2}\right)^2 - \frac{5}{4} \)
\( = \left(x + \frac{1}{2} + \frac{\sqrt{5}}{2}\right)\left(x + \frac{1}{2} - \frac{\sqrt{5}}{2}\right) \)

Hence, we have:

\( P(x) = (3x + 5) \left( x + \frac{1}{2} + \frac{\sqrt{5}}{2} \right) \left( x + \frac{1}{2} - \frac{\sqrt{5}}{2} \right) \)

Exercise &&1&& (&&1&& Question)

Let \( P(x) = x^4 - 3x^3 - 2 \). Find:

a) \( P(2) \)

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Exercise &&2&& (&&1&& Question)

Let \( P(x) = 2x^4 - x^3 + 2cx + 6 \). If \( P(1) = 21 \), find the value of \( c \).

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Exercise &&3&& (&&1&& Question)

Let \( P(x) = x^3 - 6x + 3 \) and \( Q(x) = x^2 - 3x + 1 \). Find:\( P(x) + Q(x) \)

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Exercise &&4&& (&&1&& Question)

The polynomial \(P(x) = x^3 + 3x^2 + 2x + 1\) can be written in the form \((x - 2)(x^2 + bx + c) + r\) where \(b\), \(c\), and \(r\) are real numbers. Find the values of \(b\), \(c\), and \(r\).

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Exercise &&5&& (&&1&& Question)

Find the remainder when \( P(x) = 3x^3 + 2x^2 + x + 1 \) is divided by \( 2x + 1 \).

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Exercise &&6&& (&&1&& Question)

Given that \( x + 1 \) and \( x - 2 \) are factors of \( 6x^4 - x^3 + ax^2 - 6x + b \), find the values of \( a \) and \( b \).

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Exercise &&7&& (&&1&& Question)

Factorize the Polynomial \(8x^3 + 64\).

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