# Mathematics Extension 2 – Polynomials

Polynomials

This topic extends the basis knowledge built in the Extension 1 course to encompass complex numbers. We shall break this topic into sections for study.

## Polynomials with Integer Roots

Here, the most important theorem to remember is the integer root theorem. The theorem is stated as

If a polynomial with integer coefficients has an integer root, then the root is a factor of the constant term.

We shall not prove this theorem, but shall consider examples to illustrate use of this theorem.

### Example 1

Find all the zeros of the polynomial $Latex formula$.

### Solution 1

Here, we firstly must find any roots. It is customary to check $Latex formula$ initially. It happens that:

$Latex formula$

and hence we have that $Latex formula$ is a factor of the polynomial (by the factor theorem).

That is, after performing long division (or simply by inspection) we obtain,

$Latex formula$

Now, upon factoring the quadratic we have,

$Latex formula$

Hence we have that the zeros of the polynomial are given by $Latex formula$.

### Example 2

Factorise the polynomial $Latex formula$ over the real numbers.

### Solution 2

To factorise successfully there are two methods we may use here; either we factorise using the method of grouping, or we test factors of the constant term. We shall try the latter to illustrate the techniques involved.

So, we test $Latex formula$. We find that:

$Latex formula$

Hence we have that $Latex formula$ is a factor by the factor theorem.

Performing long division, and factorising the polynomial gives:

$Latex formula$

That is,

$Latex formula$

over the $Latex formula$ numbers.

Note: At times, it is recommended that a similar theorem regarding rational roots be memorised as well. We do not recommend this, as this theorem is not stated in the syllabus. Only cases involving integer roots may be examined.

## Fundamental Theorem of Algebra

This theorem simply states that:

A polynomial of degree $Latex formula$ will have exactly zeros over the Complex field allowing for multiplicities.

This theorem is of immense importance in this topic, and should b committed to memory (verbatim).

Now, the importance of this theorem does not lie in its ability to solve equations. In fact, in that regard it is absolutely useless. The importance lies in its ability to predict the number of roots of the polynomial. This proof of existence turns out to be quite important even though the theorem gives no indication of how to find the roots.

### Example 3

Find all the roots of the polynomial $Latex formula$.

### Solution 3

We know by the fundamental theorem of algebra, that there must exist 4 roots for this Quartic polynomial, allowing for multiplicities. We hence have that:

$Latex formula$

Now, we treat the polynomial as a quadratic in terms of $Latex formula$. Factorising gives:

<=>$Latex formula$

Now, solving the equations $Latex formula$ and $Latex formula$ gives us,

$Latex formula$

Hence we have that the roots of the Quartic are $Latex formula$ only (We know there exist no other roots by the fundamental theorem of algebra).

## Complex Conjugate Root Theorem

This theorem simply states that;

If $Latex formula$ is a complex (non-real) root of the polynomial $Latex formula$ then the complex conjugate of $Latex formula$, is also a root of $Latex formula$.

We shall investigate a proof of this theorem as it is required knowledge by the syllabus.

 Theorem:If $Latex formula$ is a complex (non-real) root of the polynomial $Latex formula$ where $Latex formula$ has real coefficients, then the complex conjugate of $Latex formula$, is also a root of $Latex formula$ .Proof:Let $Latex formula$ be a non- real root of a real polynomial $Latex formula$. Then we have $Latex formula$ Taking complex conjugates of both sides gives; $Latex formula$ But, $Latex formula$ and $Latex formula$ since $Latex formula$ is real. Thus, using the fact that $Latex formula$ we now have $Latex formula$ But $Latex formula$. Thus; $Latex formula$ But the expression on the left is simply $Latex formula$. Thus $Latex formula$ That is, $Latex formula$ is also a root of $Latex formula$.

Students should memorise the theorem and proof shown here. We shall now consider an example which illustrates the use of this theorem.

### Example 4

Given that $Latex formula$ is a root of the equation latex]P(z)=2z^{3}-az^{2}+bz-2[/latex] where $Latex formula$ and $Latex formula$ are real:

a) Find the values of $Latex formula$ and $Latex formula$.

b) Hence, factorise $Latex formula$ over the real numbers.

### Solution 4

a) So we simply make the substitution into the equation, upon which we equate real parts and imaginary parts. i.e.

$Latex formula$

Upon expanding we obtain,

$Latex formula$

That is,

$Latex formula$

Equating real and imaginary parts gives:

$Latex formula$

That is,

$Latex formula$

b) To factorise $Latex formula$ over the real numbers, we simply note that since $Latex formula$ is a root of the polynomial with real coefficients then it follows that the complex conjugate of $Latex formula$, $Latex formula$ is also a root of the polynomial. Hence we have that the two of the roots of the polynomial are given by $Latex formula$ and $Latex formula$. Hence upon consideration of the sum of the roots one at a time, we have that the third root of the cubic, $Latex formula$ is given by,

$Latex formula$

That is, we have that:

$Latex formula$

However this is over the complex field.

To obtain the factorisation over the reals, we consider the product of the two complex factors.

$Latex formula$

=> $Latex formula$

$Latex formula$

Which is the factorisation over the real field.

Note: It should be noted that the product of $Latex formula$ is always real. This is because, $Latex formula$. Obviously $Latex formula$ which is real, and $Latex formula$ which is real. Hence we have that the product of two conjugate factors gives a real quadratic factor. Consequently, we can deduce that all real polynomials may be expressed as the product of real linear and quadratic factors, since all complex roots occur in pairs for complex roots of a real polynomial, and since the product of the two conjugate factors equals to a real quadratic polynomial.

## Multiple Root Theorem

The multiple root theorem simply states that;

If $Latex formula$ has $Latex formula$ where $Latex formula$ as a root of $Latex formula$multiplicity, then $Latex formula$ has $Latex formula$ as a root of multiplicity $Latex formula$.

This theorem is easily proved, and both the theorem and proof should be memorised. Below is a proof.

### Theorem

If $Latex formula$\textit{ has }$Latex formula$ where $Latex formula$ as a root of multiplicity $Latex formula$ , then $Latex formula$ has $Latex formula$ as a root of multiplicity $Latex formula$.

### Proof

Let $Latex formula$ be a polynomial given by

$Latex formula$

Where $Latex formula$ is a root of multiplicity $Latex formula$ and $Latex formula$ is some polynomial. Then, by the product rule we have

$Latex formula$

$Latex formula$

$Latex formula$

Where $Latex formula$ is some polynomial. Thus, $Latex formula$ has $Latex formula$ as a root of multiplicity $Latex formula$.

Here are some commonly asked questions regarding his theorem.

### Example 5

$Latex formula$ has a root of multiplicity $Latex formula$, find all the roots of $Latex formula$.

### Solution 5

So, if $Latex formula$ has a root of multiplicity $Latex formula$, then $Latex formula$ has a root of multiplicity $Latex formula$. Differentiating $Latex formula$ and solving the subsequent equation formed upon setting $Latex formula$ gives;

$Latex formula$

=> $Latex formula$

Now, substituting these both back into the original equation shows that $Latex formula$. Hence we have that $Latex formula$ is a double root by the multiple roots theorem. Now, using the sum of the roots one at a time gives:

$Latex formula$

Hence we have that the roots of $Latex formula$ are given by $Latex formula$.

### Example 6

If the given polynomial $Latex formula$ has a multiple root at $Latex formula$, show that

$Latex formula$

### Solution 6

It is given that the root $Latex formula$ is a multiple root but the multiplicity is not given. Thus we can only assume that it is the root of the first derivative as well.

Let $Latex formula$.

Thus, $Latex formula$ for the multiple root.

$Latex formula$

i.e.

$Latex formula$

Now, $Latex formula$ since $Latex formula$ is a root of $Latex formula$.

Thus,

$Latex formula$

But,

$Latex formula$

$Latex formula$

## Roots and Coefficients

We shall firstly introduce a table giving the relationship between the coefficients of each polynomial, and the roots.

 Order of Polynomial $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$

This notion was first introduced in the Extension 1 Mathematics topic Polynomials. The pattern is very easy to see, and students should commit this to memory.

Now, questions involving roots and coefficients are generally difficult, although they may be occasionally easier. We shall study an example to illustrate use of these techniques.

### Example 7

It is known that two of the four roots of the equation $Latex formula$ are equal but opposite in sign (i.e. $Latex formula$ ). Show that the other two roots are of the form $Latex formula$ and $Latex formula$.

### Solution 7

Let the roots be $Latex formula$. We have that:

$Latex formula$

We also have that:

$Latex formula$

Hence we have,

$Latex formula$

Now, we also have that:

$Latex formula$

Hence we have that

$Latex formula$

Hence we have two simultaneous equations to solve, namely;

$Latex formula$

$Latex formula$

Now, rearranging $Latex formula$ and substituting into $Latex formula$ gives:

$Latex formula$

=> $Latex formula$

$Latex formula$

Hence we have that

$Latex formula$

Without the loss of generality.

Hence we have that the other two roots, $Latex formula$ and $Latex formula$ are of the form $Latex formula$ where

$Latex formula$

## Finding a Polynomial given its Roots

To find the equation of a polynomial given its roots is relatively simple. If given the roots directly, then simply create a product of the factors and expand. If given the roots in terms of the roots of another polynomial, then we simply apply the algebraic trick of equating the roots of the polynomial to be found to some variable $Latex formula$ say, and then simply find the value of the original root in terms of the variable $Latex formula$, then simply substitute and rearrange to obtain in polynomial form. This trick is illustrated in the next examples. Bear in mind however that you specifically may be asked to find the equation of a polynomial using the sum and product of the roots. In this case, the algebraic trick that we apply here will not obtain you any marks.

### Example 8

Given that the equation $Latex formula$ has roots $Latex formula$, find the polynomial equation with roots $Latex formula$.

### Solution 8

In this method we shall use the algebraic substitution.

Let $Latex formula$

$Latex formula$

Now, substituting the value of $Latex formula$ into the original equation gives:

$Latex formula$

$Latex formula$

$Latex formula$ has roots $Latex formula$.

(We must express the equation in polynomial form).

### Example 9

Find the equation with roots $Latex formula$ given that the equation $Latex formula$ has roots $Latex formula$,in polynomial form.

### Solution 9

Let $Latex formula$

Inserting this value of $Latex formula$ into the given equation gives:

$Latex formula$

$Latex formula$

The question requires the equation to be in polynomial form.

Thus,

$Latex formula$

Squaring both sides gives:

$Latex formula$

$Latex formula$

$Latex formula$ Expressing the answer in polynomial form.

### Example 10

Given that $Latex formula$ has roots $Latex formula$, find the polynomial equation with roots, $Latex formula$.

### Solution 10

Let $Latex formula$.

$Latex formula$

Inserting $Latex formula$ into the given equation gives:

$Latex formula$

$Latex formula$

$Latex formula$

Expressing the equation in polynomial form.

## Finding Expressions involving the Roots of a Polynomial

In this section, we are essentially required to find the value of certain expressions given the roots of a polynomial. To illustrate the notions involved in this section, we shall answer an example question.

### Example 11

The cubic equation $Latex formula$ has 3 real non zero roots $Latex formula$. Find in terms of the constants $Latex formula$ and $Latex formula$, the values of:

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

### Solution 11

a) Now, to find the value of this we consider the following;

$Latex formula$

Using the relationship between the rots and coefficients we have that:

$Latex formula$

b) To find the next value, we must consider the following “trick”. Note that this “trick” is quite standard and must be known by every student.

Since $Latex formula$ is a root of the polynomial equation, then we have that:

$Latex formula$

The same condition holds for $Latex formula$ and $Latex formula$:

$Latex formula$

$Latex formula$

Now, performing $Latex formula$ gives:

$Latex formula$

=> $Latex formula$

c) We now perform the operation $Latex formula$ which gives:

$Latex formula$

=> $Latex formula$

In general, questions involving this subtopic are very similarly answered. The only formula to be memorised is the expansion for the sum of the squares of the roots. That is,

$Latex formula$]\alpha ^{2}+\beta ^{2}+\gamma ^{2}=(\alpha +\beta +\gamma )^{2}-2(\alpha \beta +\alpha \gamma +\beta \gamma )[/latex]

We shall now look at a further example to further reinforce technique.

### Example 12

Given that the equation $Latex formula$ has roots $Latex formula$, find:

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

d) $Latex formula$

e) $Latex formula$

### Solution 12

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

$Latex formula$

d) $Latex formula$is best found by the following method,

$Latex formula$

Now, $Latex formula$, are solutions of the above equation, thus,

$Latex formula$ (1)

$Latex formula$(2)

$Latex formula$ (3)

Now, (1) $Latex formula$ (2) $Latex formula$ (3) gives:
$Latex formula$

Thus, $Latex formula$

$Latex formula$

e) For this we shall use the previous expression.

$Latex formula$

$Latex formula$ (Multiplying both sides by $Latex formula$)

Thus, inserting $Latex formula$, into this expression gives:

$Latex formula$ (1)

$Latex formula$ (2)

$Latex formula$ (3)

Now, (1) $Latex formula$ (2) $Latex formula$ (3) gives:

$Latex formula$

$Latex formula$

$Latex formula$

## Solving Polynomials

This is the most important aspect of the polynomials topic, due to the high number of marks given in such questions. The best method to illustrate the techniques involved in solving polynomials, is to solve examples. To ensure understanding of this topic, consider the below examples.

Note that the questions in this section involve parts of the complex number topic, and it is for this reason that the mark values are high.

### Example 13

Factorise the polynomial $Latex formula$ over $Latex formula$.

### Solution 13

With polynomials that have symmetry about their coefficients, there is a set way of solving these equations. This method applies to equations of the form $Latex formula$.

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

Now, $Latex formula$ $Latex formula$,

$Latex formula$

$Latex formula$

Thus, $Latex formula$

But the question asks for the factorisation to be over $Latex formula$.

$Latex formula$ over $Latex formula$.

### Example 14

a) By considering 2 different expansions for $Latex formula$ find an expression for $Latex formula$ in terms of $Latex formula$.

b) Hence solve the equation $Latex formula$.

c) By solving the equation above using another method, find surdic expressions for $Latex formula$ and $Latex formula$.

### Solution 14

a) By De Moivre’s theorem, $Latex formula$.

By the Binomial theorem, $Latex formula$

Now, equating real and imaginary parts gives:

$Latex formula$ and $Latex formula$

Now:

$Latex formula$

Thus:

$Latex formula$

Dividing top and bottom by $Latex formula$ gives:

$Latex formula$

b) Now, we must rearrange the equation $Latex formula$ to obtain it in some form similar to the
above expression for $Latex formula$.

$Latex formula$

$Latex formula$

Now, let $Latex formula$

$Latex formula$

Thus, $Latex formula$

$Latex formula$

Thus, $Latex formula$

c) Now, the equation $Latex formula$ can be solved by the method of grouping.

$Latex formula$

$Latex formula$

$Latex formula$

Now, since $Latex formula$, then $Latex formula$ and $Latex formula$.

### Example 15

$Latex formula$, where $Latex formula$ is an integer. $Latex formula$ has two rational roots which are opposites of each other, and two non-real roots.

a) If $Latex formula$ is a non-real root of $Latex formula$, show that $Latex formula$ and $Latex formula$.

b) If the rational roots are $Latex formula$, deduce that $Latex formula$.

c) Find the rational roots and the value of $Latex formula$.

### Solution 15

a) $Latex formula$ is a Quartic polynomial and hence has four roots only by the Fundamental Theorem of Algebra. Now, let the two rational roots be $Latex formula$ and $Latex formula$, and the complex root be $Latex formula$. Since the coefficients of the polynomial are real, and since $Latex formula$ is a non-real root, then it follows that the complex conjugate of $Latex formula$, is also a root of the polynomial by the complex conjugate complex conjugate theorem.

That is, we have that the roots of the polynomial are of the form $Latex formula$. Now, consider the sum of the roots one at a time. That is, consider:

$Latex formula$

That is, we have that:

$Latex formula$

Now, $Latex formula$. Hence we have that:

$Latex formula$

=> $Latex formula$

Now, since $Latex formula$, then it follows that since:

$Latex formula$

=> $Latex formula$

Now, since we have that $Latex formula$ is non-real, then it follows that $Latex formula$ and consequently that:

$Latex formula$

b) Consider the product of the roots:

$Latex formula$

$Latex formula$

$Latex formula$

Now:

$Latex formula$

from the result in (a).

So we have:

$Latex formula$

c) So we must consider the product of the roots and the sum of the roots two at a time,

$Latex formula$

Hence we have that:

$Latex formula$

Also, from (b) we have:

$Latex formula$

Now, we let $Latex formula$ and $Latex formula$. We then have that:

$Latex formula$

and

$Latex formula$

Solving simultaneously gives:

$Latex formula$

=> $Latex formula$

=> $Latex formula$

Now, $Latex formula$ and hence we have that

$Latex formula$

We then have that:

$Latex formula$

That is, the rational roots are $Latex formula$ and $Latex formula$.

Substituting $Latex formula$ into $Latex formula$and then solving for $Latex formula$ gives:

$Latex formula$

=> $Latex formula$

=> $Latex formula$

## Finding the Number of Real Roots of a Polynomial

At times we do not wish to solve for the roots of a polynomial, but instead require the number of real roots the polynomial possesses. There is a method of doing so and its use involves consideration of the product of the $Latex formula$-values of the stationary points. This method is mainly applied to cubic polynomials, however the results may be extended to include special cases of other polynomials. Consider the below diagrams:

Students should note that no memorisation is required. When required, simply sketch the required situation, after which the other results follow. Consider the below example which illustrates use of this powerful technique.

### Example 16

Consider the function $Latex formula$, where $Latex formula$ and $Latex formula$ are positive. Show that the function $Latex formula$ has 3 distinct real roots if and only if $Latex formula$.

### Solution 16

For the function to have three distinct real roots, we must have that the product of the $Latex formula$-values of the stationary points must be less than $Latex formula$.

So, firstly we differentiate the function and find the stationary points.

$Latex formula$

For stationary points.

=> $Latex formula$

$Latex formula$ and $Latex formula$. Hence we have that the stationary points are $Latex formula$ and $Latex formula$. Now, the product of the $Latex formula$-values of the stationary points must be less than zero for polynomial to have three distinct real roots. That is,

$Latex formula$

$Latex formula$

Now, $Latex formula$ is positive as is $Latex formula$. Hence we have that:

$Latex formula$

$Latex formula$

Hence there exists three distinct real roots for $Latex formula$ if and only if $Latex formula$.

## Partial Fractions

The method of partial fractions is a pivotal part of your mathematical education. This technique essentially involves expressing a rational function (polynomial divided by polynomial) as the sum of two or more other rational functions. The use of such a technique becomes apparent in the topic Integration in which rational functions are integrated using this method.

The method of partial fractions essentially involves the equating of the rational function to the required sum. Now, in order to do so, we must set the correct numerator on each expression. In general, when placing a numerator on a linear expression, place a constant $Latex formula$etc. $Latex formula$, and when placing a numerator on a quadratic place a linear expression $Latex formula$. Be sure that the same constant is not used twice. Now, these are the only two cases required. Upon doing so, we simply consider the sum and rearrange to find the value of the constants by making substitutions and equating coefficients. We shall illustrate these in the below examples.

Note: We do not need to consider any other case (other than quadratic and linear denominators) since all real polynomials may be expressed as the product of real linear and quadratic expressions.

Now before we move on to examples, be careful to note that the degree of the numerator must be less than the degree of the denominator before undergoing the partial fraction decomposition. To solve this problem, we simply perform the long division, and express the polynomial as the sum of a quotient and a remainder over the divisor polynomial. The below examples illustrate this method.

### Example 17

Express the rational function:

$Latex formula$

In the form:

$Latex formula$

where $Latex formula$.

### Solution 17

Firstly, we notice that the denominator is $Latex formula$. Using this fact, we may express $Latex formula$ in a more useful form

$Latex formula$

Which is the required form.

Note: In general questions do not involve prior preparation of a rational function for the use of the method of partial fractions.

We shall now consider example questions which illustrate the method involved in obtaining the partial fraction expansion.

### Example 18

Express the following rational functions as the sum of partial fractions;

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

### Solution 18

a) As stated above we simply equate the expression to the required expression, whilst placing constants for the values in the numerator (since both denominators are linear).

$Latex formula$

Now, multiplying through by $Latex formula$ gives:

$Latex formula$

Now, letting $Latex formula$ gives:

$Latex formula$

Letting $Latex formula$ gives:

$Latex formula$

Hence we have that:

$Latex formula$

b) Here we equate the rational function to the required final partial fraction expression, taking care to place a linear factor for the numerator on the quadratic expression.

$Latex formula$

Multiplying through by $Latex formula$ gives:

$Latex formula$

Now, let $Latex formula$. We thus have that:

$Latex formula$

Now, we equate the coefficient of $Latex formula$ on the LHS and the RHS. Doing so gives:

$Latex formula$

Now, $Latex formula$. Hence we have that:

$Latex formula$

Now, equating the coefficients of the constant term on LHS and RHS gives:

$Latex formula$

Hence we have that:

$Latex formula$

Hence we have that:

$Latex formula$

Note: The use of the method of equating coefficients and substitution is usually faster than simply using a single method.

c) Here we perform the same sequence of steps as before:

$Latex formula$

Now, multiplying through by $Latex formula$ gives:

$Latex formula$

Now, the method of substitution proves to be difficult here. Instead, we shall equate coefficients.

Coefficient of $Latex formula$:

$Latex formula$

Coefficient of $Latex formula$:

$Latex formula$

Coefficient of $Latex formula$:

$Latex formula$

Coefficient of constant:

$Latex formula$

Now solving $Latex formula$ and $Latex formula$ simultaneously gives:

$Latex formula$

Solving $Latex formula$ and $Latex formula$ simultaneously gives:

$Latex formula$

Hence we have that the partial fraction expansion is given by:

$Latex formula$

Almost every question falls into a similar category to one of the three questions solved above. Be sure to understand the methods involved. Partial fraction expansions involving squares in the linear terms are not included in the syllabus, however they are asked in HSC questions. Students are not expected to be able to predict the final partial fractions expansion, and are instead given the expression required and students are only required to find the values of the constants.

Note: In the above examples, the use of the method of substitution may seem contradictory. We are substituting a point into the expression, although the expression is derived from another expression within which we are unable to substitute those points (due to a division by zero). We are in actual fact not making a direct substitution, but instead taking a limit as $Latex formula$ approaches the required value. However, this produces the same value as if we had simply substituted the point in, and hence we skip the rigor of the method, and simply make the substitution rather than take the limit every time we use the technique.