AOS1 Topic 4: Polynomial Functions

Basic Definitions

1. 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 $.

2. Leading Term

The leading term of a polynomial $P(x)$, denoted as $a_n x^n$, is the term with the highest degree (i.e., the highest exponent) among all terms with non-zero coefficients.

3. Degree of a Polynomial

The degree of a polynomial is the highest exponent $n$ in the polynomial. For example, the polynomial $P(x)=4x^3+3x^2−2x+5$ has a degree of $3$.

4. Constant Term

The constant term is the term of degree 0 in a polynomial, denoted as $a_0$. It is the term not involving the variable $x$. For example, in $P(x)=2x^3+5x−7$, the constant term is $−7$.

Types of Polynomial Functions

Quadratic Function

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

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

Example:

\[f(x)=3x^2−5x+2\]

Cubic Function

A cubic function is a polynomial function of degree 3. The general form is:

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

Example:

\[f(x)=2x^3−4x^2+x−5\]

Quartic Function

A quartic function is a polynomial function of degree 4. The general form is:

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

Example:

\[f(x)=x^4−3x^3+2x^2−x+4\]

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:

1. Addition

The sum of two polynomials is obtained by adding their corresponding coefficients.

\[ P(x) + Q(x) = (x^3 + 3x^2 + 2) + (2x^2 + 4) \]

\[ = x^3 + 5x^2 + 6 \]

2. Subtraction

The difference of two polynomials is obtained by subtracting their corresponding coefficients.

\[ P(x) - Q(x) = (x^3 + 3x^2 + 2) - (2x^2 + 4) \]

\[ = x^3 + x^2 - 2 \]

3. Multiplication

The product of two polynomials is obtained by multiplying each term of one polynomial by each term of the other polynomial and combining like terms.

\[ P(x)Q(x) = (x^3 + 3x^2 + 2)(2x^2 + 4) \]

\[ = (x^3 + 3x^2 + 2) \times 2x^2 + (x^3 + 3x^2 + 2) \times 4 \]

\[ = 2x^5 + 6x^4 + 4x^3 + 4x^3 + 12x^2 + 8 \]

\[ = 2x^5 + 6x^4 + 4x^3 + 16x^2 + 8 \]

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:

\[\deg(f + g) \leq \max ( \deg(f), \deg(g) )\]

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

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.

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:

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

Theorems Related to Polynomials

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) $

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.

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 $.

Example:

Let $P(x)=x^3−4x^2+5x−2$

To check if $x−1$ is a factor, find $P(1)$:

\[P(1)=1−4+5−2=0\]

Since $P(1)=0$, $x−1$ is a factor of $P(x)$.

Substituting $P(2)$ gives

\[P(2)=8−16+10−2=0\]

Therefore $x-2$ is also a factor.

\[P(x) = (x-2)(x-1)^2\]

Try with random polynomials and verify your answer with the simulation below.

Created with GeoGebra ® , by Steve Phelps, Link

The Rational Root Theorem

If $P(x)=a_nx^n+a_{n−1}x^{n−1} + \dots + a_1 x + a_0$ is a polynomial with integer coefficients, then any possible rational root $\frac{p}{q}$ satisfies:

$p$ divides the constant term $a_0$,

$q$ divides the leading coefficient $a_n$.

Example:

Consider the cubic polynomial

\[ P(x) = 2x^3 - x^2 - x - 3 \]

If the equation $ P(x) = 0 $ has a solution $ \alpha$ that is an integer, then $ \alpha$ 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.

The rational-root theorem helps us with this. It says that if \alpha and $\beta$ have highest common factor 1 and if $\beta x + \alpha$ is a factor of $ 2x^3 - x^2 - x - 3 $, then $\beta$ divides 2 and \alpha divides -3.

Therefore, if $ -\frac{\alpha}{\beta} $ is a solution of the equation $ P(x) = 0 $, then $\beta$ must divide 2 (so $\beta$ has to be either 1 or 2) and \alpha must divide -3 (so $\alpha$ must be $\pm 1$ or $\pm 3$ . So the only value of $\beta$ that needs to be considered is 2, and $ \alpha = \pm 3 $ or $ \alpha = \pm 1 $.

We can test these through the factor theorem. That is, check $ P(\pm \frac{1}{2}) $ and $ P( \pm \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 .

Common Misconceptions

Misconception 1: The Degree of a Polynomial Equals the Number of Terms

Clarification: The degree of a polynomial is the highest exponent of the variable, not the number of terms. For example, $P(x)=x^5+2x+1$ has a degree of $5$, while having three terms. Just focus on the powers of $x$ when finding the degree.

Misconception 2: All Polynomials Have Real Roots

Clarification: Not all polynomials have real roots. Depending on the discriminant and the polynomial's degree, a polynomial may have real or complex roots. For example, $P(x)= x^2 + 1$ has no real roots. So if you can't calculate a root, it just be because it doesn't have one.

Misconception 3: If $P( \alpha)=0$, Then $x+ \alpha$ is a Factor of $P(x)$

Clarification: If $P(\alpha) = 0$, then $x− \alpha$ is a factor of $P(x)$, not $x+ \alpha$.

Summary: Polynomial Functions

Polynomial Function: A function of the form$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $n$ is a non-negative integer and $a_n \neq 0$.

Key Terms:

Leading Term: The term with the highest degree ($a_n x^n$).
Degree: The highest exponent in the polynomial.
Constant Term: The term of degree 0 ($a_0$).

Arithmetic of Polynomials:

Addition/Subtraction: Combine like terms by adding or subtracting their coefficients.

Multiplication: Distribute each term of one polynomial across the other and combine like terms.

Division and Factorisation:

Polynomial Long Division: Similar to normal long division, used to divide polynomials.

Remainder Theorem: The remainder of $P(x)$ divided by $x− \alpha$ is $P(\alpha)$.

Factor Theorem: $x - \alpha$ is a factor of $P(x)$ if and only if $P(\alpha) = 0.

Rational-Root Theorem: Provides possible rational roots based on factors of the constant and leading coefficients.

Sums and Differences of Cubes:

Sum of Cubes: $x^3 + a^3 = (x + a)(x^2 - ax + a^2)$.

Difference of Cubes: $x^3 - a^3 = (x - a)(x^2 + ax + a^2)$.

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

Practice Makes Perfect

Exercise 2

Practice Makes Perfect

Exercise 3

Practice Makes Perfect

Exercise 4

Practice Makes Perfect

Exercise 5

Practice Makes Perfect

Exercise 6

Practice Makes Perfect

Exercise 7

Practice Makes Perfect

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