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
Contents
 1 Polynomials with Integer Roots
 2 Fundamental Theorem of Algebra
 3 Complex Conjugate Root Theorem
 4 Multiple Root Theorem
 5 Roots and Coefficients
 6 Finding a Polynomial given its Roots
 7 Finding Expressions involving the Roots of a Polynomial
 8 Solving Polynomials
 9 Finding the Number of Real Roots of a Polynomial
 10 Partial Fractions
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 1Find all the zeros of the polynomial . Solution 1Here, we firstly must find any roots. It is customary to check initially. It happens that: and hence we have that is a factor of the polynomial (by the factor theorem). That is, after performing long division (or simply by inspection) we obtain, Now, upon factoring the quadratic we have, Hence we have that the zeros of the polynomial are given by . 
Example 2Factorise the polynomial over the real numbers. Solution 2To 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 . We find that: Hence we have that is a factor by the factor theorem. Performing long division, and factorising the polynomial gives: That is, over the 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 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 3Find all the roots of the polynomial . Solution 3We know by the fundamental theorem of algebra, that there must exist 4 roots for this Quartic polynomial, allowing for multiplicities. We hence have that: Now, we treat the polynomial as a quadratic in terms of . Factorising gives: <=> Now, solving the equations and gives us, Hence we have that the roots of the Quartic are only (We know there exist no other roots by the fundamental theorem of algebra). 
Complex Conjugate Root Theorem
This theorem simply states that;
If is a complex (nonreal) root of the polynomial then the complex conjugate of , is also a root of .
We shall investigate a proof of this theorem as it is required knowledge by the syllabus.
Theorem:If is a complex (nonreal) root of the polynomial where has real coefficients, then the complex conjugate of , is also a root of .Proof:Let be a non real root of a real polynomial . Then we have
Taking complex conjugates of both sides gives; But, and since is real. Thus, using the fact that we now have But . Thus; But the expression on the left is simply . Thus That is, is also a root of . 
Students should memorise the theorem and proof shown here. We shall now consider an example which illustrates the use of this theorem.
Example 4Given that is a root of the equation latex]P(z)=2z^{3}az^{2}+bz2[/latex] where and are real: a) Find the values of and . b) Hence, factorise over the real numbers. Solution 4a) So we simply make the substitution into the equation, upon which we equate real parts and imaginary parts. i.e. Upon expanding we obtain, That is, Equating real and imaginary parts gives: That is, b) To factorise over the real numbers, we simply note that since is a root of the polynomial with real coefficients then it follows that the complex conjugate of , is also a root of the polynomial. Hence we have that the two of the roots of the polynomial are given by and . Hence upon consideration of the sum of the roots one at a time, we have that the third root of the cubic, is given by, That is, we have that: However this is over the complex field. To obtain the factorisation over the reals, we consider the product of the two complex factors. => Which is the factorisation over the real field. 
Note: It should be noted that the product of is always real. This is because, . Obviously which is real, and 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 has where as a root of multiplicity, then has as a root of multiplicity .
This theorem is easily proved, and both the theorem and proof should be memorised. Below is a proof.
TheoremIf \textit{ has } where as a root of multiplicity , then has as a root of multiplicity . ProofLet be a polynomial given by Where is a root of multiplicity and is some polynomial. Then, by the product rule we have Where is some polynomial. Thus, has as a root of multiplicity . 
Here are some commonly asked questions regarding his theorem.
Example 5has a root of multiplicity , find all the roots of . Solution 5So, if has a root of multiplicity , then has a root of multiplicity . Differentiating and solving the subsequent equation formed upon setting gives; => Now, substituting these both back into the original equation shows that . Hence we have that is a double root by the multiple roots theorem. Now, using the sum of the roots one at a time gives: Hence we have that the roots of are given by . 
Example 6If the given polynomial has a multiple root at , show that Solution 6It is given that the root 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 . Thus, for the multiple root. i.e. Now, since is a root of . Thus, But, 
Roots and Coefficients
We shall firstly introduce a table giving the relationship between the coefficients of each polynomial, and the roots.
Order of Polynomial  
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 7It is known that two of the four roots of the equation are equal but opposite in sign (i.e. ). Show that the other two roots are of the form and . Solution 7Let the roots be . We have that: We also have that: Hence we have, Now, we also have that: Hence we have that Hence we have two simultaneous equations to solve, namely; Now, rearranging and substituting into gives: => Using the quadratic formula, Hence we have that Without the loss of generality. Hence we have that the other two roots, and are of the form where 
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 say, and then simply find the value of the original root in terms of the variable , 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 8Given that the equation has roots , find the polynomial equation with roots . Solution 8In this method we shall use the algebraic substitution. Let Now, substituting the value of into the original equation gives:
has roots . (We must express the equation in polynomial form). 
Example 9Find the equation with roots given that the equation has roots ,in polynomial form. Solution 9Let Inserting this value of into the given equation gives:
The question requires the equation to be in polynomial form. Thus,
Squaring both sides gives:
Expressing the answer in polynomial form. 
Example 10Given that has roots , find the polynomial equation with roots, . Solution 10Let . Inserting into the given equation gives:
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 11The cubic equation has 3 real non zero roots . Find in terms of the constants and , the values of: a) b) c) Solution 11a) Now, to find the value of this we consider the following; Using the relationship between the rots and coefficients we have that: 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 is a root of the polynomial equation, then we have that: The same condition holds for and : Now, performing gives: => c) We now perform the operation which gives: => 
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,
]\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 12Given that the equation has roots , find: a) b) c) d) e) Solution 12a) b) c) d) is best found by the following method, Now, , are solutions of the above equation, thus, (1) (2) (3) Now, (1) (2) (3) gives: Thus, e) For this we shall use the previous expression.
(Multiplying both sides by ) Thus, inserting , into this expression gives: (1) (2) (3) Now, (1) (2) (3) gives:

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 13Factorise the polynomial over . Solution 13With polynomials that have symmetry about their coefficients, there is a set way of solving these equations. This method applies to equations of the form . Now, , Thus, But the question asks for the factorisation to be over . over . 
Example 14a) By considering 2 different expansions for find an expression for in terms of . b) Hence solve the equation . c) By solving the equation above using another method, find surdic expressions for and . Solution 14a) By De Moivre’s theorem, . By the Binomial theorem, Now, equating real and imaginary parts gives: and Now: Thus: Dividing top and bottom by gives: b) Now, we must rearrange the equation to obtain it in some form similar to the Now, let Thus,
Thus, c) Now, the equation can be solved by the method of grouping.
Now, since , then and . 
Example 15, where is an integer. has two rational roots which are opposites of each other, and two nonreal roots. a) If is a nonreal root of , show that and . b) If the rational roots are , deduce that . c) Find the rational roots and the value of . Solution 15a) is a Quartic polynomial and hence has four roots only by the Fundamental Theorem of Algebra. Now, let the two rational roots be and , and the complex root be . Since the coefficients of the polynomial are real, and since is a nonreal root, then it follows that the complex conjugate of , 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 . Now, consider the sum of the roots one at a time. That is, consider: That is, we have that: Now, . Hence we have that: => Now, since , then it follows that since: => Now, since we have that is nonreal, then it follows that and consequently that: b) Consider the product of the roots: Now: from the result in (a). So we have: c) So we must consider the product of the roots and the sum of the roots two at a time, Hence we have that: Also, from (b) we have: Now, we let and . We then have that: and Solving simultaneously gives: => => Now, and hence we have that We then have that: That is, the rational roots are and . Substituting into and then solving for gives: => => 
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 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 16Consider the function , where and are positive. Show that the function has 3 distinct real roots if and only if . Solution 16For the function to have three distinct real roots, we must have that the product of the values of the stationary points must be less than . So, firstly we differentiate the function and find the stationary points. For stationary points. => and . Hence we have that the stationary points are and . Now, the product of the values of the stationary points must be less than zero for polynomial to have three distinct real roots. That is, Now, is positive as is . Hence we have that: Hence there exists three distinct real roots for if and only if . 
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 etc. , and when placing a numerator on a quadratic place a linear expression . 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 17Express the rational function: In the form: where . Solution 17Firstly, we notice that the denominator is . Using this fact, we may express in a more useful form 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 18Express the following rational functions as the sum of partial fractions; a) b) c) Solution 18a) 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). Now, multiplying through by gives: Now, letting gives: Letting gives: Hence we have that: 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. Multiplying through by gives: Now, let . We thus have that: Now, we equate the coefficient of on the LHS and the RHS. Doing so gives: Now, . Hence we have that: Now, equating the coefficients of the constant term on LHS and RHS gives: Hence we have that: Hence we have that: 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: Now, multiplying through by gives: Now, the method of substitution proves to be difficult here. Instead, we shall equate coefficients. Coefficient of : Coefficient of : Coefficient of : Coefficient of constant: Now solving and simultaneously gives: Solving and simultaneously gives: Hence we have that the partial fraction expansion is given by: 
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 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.