# 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

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1 9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role of a place marker. The role of the x is to provide positions in the expression which can be replaced (substituted) by a value. The numbers a 0, a 1,... are called the coefficients of the polynomial. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11. The powers of x in a polynomial must be non-negative integers and there must be a finite number of terms. Other expressions have the appearance of being polynomials but, because some powers are negative or fractional, or because there are infinitely many terms, they are not considered to be polynomials. Example 2: The expressions x + 1 x 1 x + x 2 and 1 + x + x2 2! + x3 3! +... are not polynomials even though they are built up from powers of x. The coefficients can be rational numbers (forming the set Q), real numbers (forming the set R), or complex numbers (forming the set C). These systems, called fields, have the property that x 1 exists for every non-zero x. (This somewhat loose description will do for our present purposes. An exact definition comprises 11 separate properties, or axioms.) Sometimes we shall restrict our attention to polynomials with integer coefficients even though the integers don't form a field. Notice that a rational polynomial can always be converted to an integer polynomial by multiplying by a common denominator. Example 3: The rational polynomial x3 2 x is equal to 88x3 70x It is also possible to consider polynomials whose coefficients are integers modulo p for some prime number. These systems are fields and the theory of polynomials over these finite fields works nicely even if some of the results look a bit strange. For example over Z 2, the field of integers mod 2 (where there are just two elements, 0 and 1 with 1 + 1=0) the factorisation of x = (x + 1) 2, since the 2x term is equal to

3 1 Now it is possible to write 1 x as 1 + x + x but although this looks like a polynomial it has infinitely many terms while a polynomial, by definition, has only finitely many. An expression like 1 + x + x is called a power series. The system F[x] of polynomials over a field, in fact, behaves much more like the system of integers (where only ±1 have integer inverses). Theorem 1: For polynomials a(x), b(x) F [x], where F is a field: deg [a(x)b(x)] = deg a(x) + deg b(x) Proof: This follows from the fact that (a m x m +... )(b n x n +... ) = am b n x m+n +..., and the fact that if a m and b n are non zero then a m b n is also non-zero. Thus the degree of a product is the sum of the degrees. Does this remind you of something? The degree function behaves like the logarithm function. Theorem 2: For polynomials a(x), b(x) F[x] deg [a(x) + b(x)] max[deg a(x), deg b(x)). There is nothing really surprising in this result except for the question as to why lessthan-or-equals rather than just equals. The answer is that when we add two polynomials of the same degree the leading coefficients can cancel thereby producing a polynomial of lower degree. Example 7: If a(x) = 2x 2 x + 7 and b(x) = 2x 2 + 4x + 1 then a(x) + b(x) = 3x + 8, which has smaller degree than either a(x) or b(x) Division and Remainders As mentioned earlier, one polynomial does not usually divide exactly into another. Like the system of integers we are usually left with a remainder. We get exact divisibility precisely when the remainder is zero. Furthermore this remainder, when it is not zero, is in some sense smaller than whatever we are dividing by. For polynomials, smaller means of smaller degree. The process of obtaining the remainder on dividing one polynomial by another is very similar to the familiar long-division algorithm. Example 8: 2x + 4 x 2 2x + 7 ) 2x 3 + 5x 3 2x 3 4x 2 +14x 4x 2 9x 3 4x 2 8x + 28 x

4 From this calculation we compute the remainder on dividing 2x 3 + 5x 3 by x 2 2x + 7 to be x 31. Note that the remainder has lower degree than x 2 2x + 7, the polynomial we are dividing by. Note also how we write the terms neatly underneath others of the same degree. The result of the calculation can also be expressed as: 2x 3 + 5x 3 = (x 2 2x +7)(2x + 4) + ( x 31). Theorem 3 (Division Algorithm): If a(x), b(x) are polynomials and b(x) is non-zero then a(x) = b(x)q(x) + r(x) for some polynomials q(x) and r(x) where either r(x) = 0 or deg r(x) < deg b(x). The polynomial q(x) is called the quotient and r(x) is called the remainder. If the remainder on dividing a(x) by b(x) is zero we say that b(x) divides a(x), or that a(x) is a multiple of b(x). If we can't be bothered saying it in words we just write b(x) a(x) and read it as b(x) divides a(x). Example 9: Since x 2 5x + 6 = (x 2)(x 3) it is true that x 2 x 2 5x + 6. But x 2 2x + 7 does not divide 2x 3 + 5x Substitution and the Remainder Theorem If f(x) F[x], in other words f(x) is a polynomial in x with coefficients coming from the field F, and F we define f( ) to be the number, in F, that results from replacing, or substituting, x in the polynomial by the value. For example if f(x) = x 2 + x 2 then f(2) = = 4, f(0) = 2 and f(1)=0. The following theorem connects the ideas of substitution and remainder. Theorem 4 (Remainder Theorem): The remainder on dividing f(x) by x is f( ). Proof: By the Division Algorithm, f(x) = (x )q(x) + r(x) for some polynomials q(x), r(x) and the remainder r(x) is either zero or has degree less than 1. In other words r(x) must be a constant polynomial, so we can drop the (x) and just call it r. Now substituting x = into the equation f(x) = (x )q(x) + r, we get f( ) = r. Corollary: The polynomial f(x) is divisible by x if and only if f( ) = 0. Example 11: We saw that if f(x) = x 2 + x 2 then f(2) = 4. The Remainder Theorem concludes that 4 must be the remainder on dividing f(x) by x 2. Numbers which produce zero when substituted into a polynomial f(x) are just the solutions of the polynomial equation f(x) = 0. They are called the roots of the polynomial and they are quite important features of the polynomial. 104

5 9.6. Roots of Polynomials A root of a polynomial f(x) is a number,, so that f( )=0. Solving a polynomial equation f(x) = 0 therefore means finding all its roots. But where do we look for potential roots? From the coefficient field. But here we have to be a little careful. Does the polynomial f(x) = x have any roots? That depends. The coefficients are real numbers so we could consider f(x) as belonging to the set of real polynomials, R[x]. If so, there are no roots. But we can just as validly consider f(x) as belonging to the set of complex polynomials C[x] (remember the set C includes all of the real numbers). Then we would have two roots for f(x), namely ± i. Frequently we switch from one field to another. So we can say that x has no real roots, but two complex roots. The corollary to the Remainder Theorem can be expressed by saying that the number is a root of f( ) if and only if x is one of its factors. Now the polynomial x has degree 1 and it is called a linear polynomial. So there is a connection between linear factors and roots of a polynomial. Theorem 5: A polynomial has a zero if and only if it has a linear factor. Proof: Whenever we have a root,, we have a linear factor x. Conversely having a linear factor bx + c for a polynomial means that we have a root x = c/b If we know one root of a polynomial we can use the remainder theorem and divide by the corresponding linear factor. The other roots will then be roots of the quotient. Example 12: Given that x = 2 is a root of x 3 x 2 x 2, find the other two roots. Solution: By the remainder theorem x 2 is a factor. Dividing the cubic by x 2 we get the other factor which we proceed to solve. So x 3 x 2 x 2 = (x 2)(x 2 + x + 1). The other roots are the roots of x 2 + x + 1, viz ½( 1 ± 3i) Sum and Product of Roots If and are the roots of the quadratic ax 2 + bx + c then: + = b a (sum of the roots) and = c a (product of the roots). Example 13: If and are the roots of x 2 + 3x + 7, find: (i) + ; (ii) ; (iii) ; (iv)

6 Solution: (i) + = 3; (ii) = 7; (iii) = ( + ) 2 2 = 9 14 = 5 (if you are worried by getting a negative sum of squares just remember that all this means is that the roots are non-real) ; (iv) = + = 3 7. Example 14: If and are the roots of 2x 2 5x + 1, find (i) ; (ii) +. Solution: + = 5/2 and = ½. We must now express each expression in terms of these. (i) = ( + ) = ( + ) 3 3 ( + ) = = (ii) ( + ) 2 = = = Hence + = 2. (The quadratic does have positive real roots and we are taking positive square roots throughout.) For the roots,, of the cubic ax 3 + bx 2 + cx + d we get similar results, but there is an additional combination of the roots we need to consider. + + = b a (sum of the roots); + + = c a (sum of the roots taken two at a time); = d a (product of the roots). Example 15: If,, are the roots of the cubic x 3 + 5x 2 2x 3, find the value of Solution: = + + = 2 3. The three expressions + +, + + and are symmetric in, and. This means that, and can be permuted in any order and the value of these expressions is unchanged. It is obvious that any expression that we can build up from these three expressions must also be symmetric. What is not obvious, but is nevertheless true, is the converse. Every symmetric expression in, and can be expressed in terms of these three special ones. They are known as the elementary symmetric functions. Example 16: Express the following symmetric functions in terms of the elementary symmetric ones: (i) + + ; (ii) ; (iii) Solution: (i) + + = = ( + + ) = ( + + )2 2 ( + + ) ; (ii) = ( + + )( + + ) 3 ; 106

7 (iii) = ( + + ) = ( + + ) 3 3( ) = ( + + ) 3 3( + + )( + + ) = ( + + ) 3 3( + + )( + + ) + 3. The above results extend to polynomials of higher degree. If f(x) = a n x n + a n 1 x n a 1 x + a 0 is a polynomial of degree n with roots 1, 2,, n then 1 2 r, the sum of all products of roots taken r at a time, is ( 1) r a n r a n. Moreover every symmetric function in the i can be expressed in terms of these elementary symmetric functions. Example 17: If,,, are the roots of x 4 + 5x 3 3x 2 3x 7 find Solution: = ( ) 2 2( ) = 2 = 5 2( 3) = Real and Complex Polynomials A real polynomial can be graphed in the usual way and real roots correspond to places where the curve cuts the x-axis. A real polynomial has at least one real root if its graph crosses the x-axis. Theorem 6: Real polynomials of odd degree have at least one real root. Proof: If a(x) has odd degree, then a(x) as x and a(x) as x. It therefore takes both positive and negative values and since it moves continuously between positive and negative values it must cross the x-axis in at least one place. Real polynomials of even degree can have real roots too, but there are plenty which don't (such as x 2 + 1). We can't graph a polynomial with complex coefficients without using 4 dimensions. However a very important theorem, easy to state but not so easy to prove, states that every complex polynomial (apart from constants) has complex roots. Theorem 7 (Fundamental Theorem of Algebra): If f(x) C[x] and deg f(x) 1, then f( ) = 0 for some C. Example 17: You will know that every real quadratic ax 2 + bx + c has roots in C from the quadratic equation formula: x = b b2 4ac 2a with real roots if b² 4ac and non-real roots if b 2 < 4ac. 107

9 Theorem 8 (Conjugate Pairs): The non-real roots of a real polynomial come in conjugate pairs. Proof: Suppose is a root of the real polynomial a(x) = a n x n + an 1 x n a1 x + a 0. Then a( ) = 0. Hence a n n + an 1 n a1 + a 0 = 0. Taking conjugates of both sides we see that the conjugate of is also a root. Example 19: The five roots of f(x) = x 5 2x 4 + 6x 3 2x 2 + 5x are 0, 1 2i and i. The factorisation of f(x) over C is f(x) = x(x i)(x+i)(x 1 2i)(x+1+2i). Grouping the linear factors which correspond to conjugate roots this becomes f(x) = x(x 2 + 1)(x 2 2x + 5), a factorisation over R. root. Theorem 8 can be useful in factorising certain real polynomials if we are given a non-real Example 20: Assuming that 2 + i is a root of x 4 6x 3 22x x 175 factorise it completely over the rationals. Solution: Since the coefficients are real, 2 i is also a root and hence (x 2 i)(x 2 + i) = x 2 4x + 5 is a factor. Dividing by this quadratic we get the other factor: z 2 2x Formulae for Higher Degree Polynomials We all know how to solve a quadratic equation ax 2 + bx + c = 0. Unless the factors are obvious it is probably quicker to use the quadratic equation formula: b b 2 4ac 2a. There is a similar, but more complicated, formula for cubics. Every cubic equation can be expressed in the form x 3 + bx 2 + cx + d = 0 by dividing through by the leading coefficient. Then by substituting y = x + b 3 we can eliminate the squared term and write y as a solution to a cubic of the form y 3 + cy + d = 0. For such a cubic the three solutions for y are 3 d 2 + d c d 2 d c3 27. There are three cube roots for every non-zero complex number, so this formula appears to give 9 solutions altogether. However the cube roots have to be paired up in a special way, giving only three solutions in all. Even if the roots turn out eventually to be real the process involves working with non-real numbers. This is why, despite the misgivings mathematicians had about whether square roots of 1 really exist, they could not ignore the fact that even if they didn t exist they were really useful. You will notice that the cubic formula is of the same type as the quadratic one. You start with the coefficients and then carry out certain arithmetic operations to find the solutions. These 109

10 operations are addition, subtraction, multiplication and division and the extraction of roots. In the case of the cubic we need to take square roots as well as cube roots. There exists an exact formulae for the quartic, along the same lines. The cubic and the quartic fomulae were discovered in 1515 and 1545 respectively and, as you would expect, the search was on for a quintic equation formula. You might ask who would really want such formulae. Well, don t forget that this was before calculus, so that Newton s method, or other iterative process based on derivatives were not possible. But what is more to the point was that solving polynomial equations became a fairground stunt. A moderately educated member of the public would choose some numbers, usually integers, and construct the polynomial that has these as its zeros. The operator of the stall would wager money that he could solve the polynomial in a certain time! The quintic formula was never found in the two centuries following the discovery of the quartics formula. There had to be a solution, it was just proving to be too complicated. Or, so everyone thought. In fact the question was finally settled in the early part of the nineteenth century when the Norwegian mathematician, Abel, showed that no such formula exists for degree 5 or more. But it was Evariste Galois, a young teenager, who was killed in a duel at the age of 20, who devised a wonderfully powerful theory that determines which polynomials of degree five or more can be solved in this way. A solution involving just the four basic arithmetic operations plus the extraction of roots is called a solution by radicals. Galois invented a whole new branch of mathematics, which we call group theory and applied this to permutations of the zeros of a polynomial in a way that we now call Galois theory, to determine whether a given polynomial is soluble by radicals. All this by an unknown teenage French boy, who scribbled most of his mathematics in prison (he was a student rebel), by the age of 19. If you ever get a chance to read his story, do so. It is the most dramatic story of any mathematician in history! Solving Systems of Polynomial Equations in One Variable Example 21: Solve the system x 3 1= 0 x 4 + x 3 x 2 + x 2 = 0. In theory we can always solve the first equation and test them in the second. That would indeed work in this case, because we know that the cube roots of 1 are 1, and 2, where = e 2 i/3. But in general this would not be at all feasible. The secret is to find the greatest common divisor of the two polynomials. This will be the monic (leading coefficient 1) of largest degree that divides both polynomials. The common solutions to the two polynomial equations will simply be all the solutions of the GCD. The process of finding the GCD, though it can sometimes be a little messy, is really not that hard. We still have to solve the equation for the GCD, but it is often the case that the degree of the GCD is quite small. A greatest common divisor (or GCD) of f(x) and g(x) is a polynomial of largest degree which divides them both. 110

11 Example22: The polynomial x 1 divides both x 2 1 and x 2 2x + 1. No polynomial of higher degree divides both so x 1 is a greatest common divisor. But so are 2x 2 and ¼x ¼. What distinguishes x 1 from these others is that it is monic its leading coefficient is 1. We would like to be able to define the GCD to be the monic one. But isn t it conceivable that there is more than one? Imagine that we had two polynomials f(x) and g(x). Suppose further that both are divisible by x 2, both are divisible by x + 5, and that no quadratic, or polynomial of higher degree divides them both. If such a situation was possible, what would be the GCD. of f(x) and g(x)? Would the GCD be x 2 or would it be x + 5? Both would be common divisors of larger degree that divide both and both are monic. The answer is that such a situation cannot arise. If x 2 and x + 5 were both common divisors, their product (x 2)(x + 5) would also have to be a common divisor. But the product of two divisors is not always a divisor. For example both x 2 1 and x 2 2x + 1 are divisors of the cubic (x 1) 3 (x + 1) but their product, (x 2 1)(x 2 2x + 1), has degree 4 and so can t possibly be a divisor of the cubic. To help clarify this situation we show that there is another characterisation of GCDs Theorem 9: Let S be the set of all polynomials which are expressible in the form: a(x)h(x) + b(x)k(x). Then any polynomial in S of lowest degree is a greatest common divisor of a(x) and b(x). Proof: S includes, of course, the zero polynomial, where h(x) = k(x) = 0, and it includes both a(x) and b(x) themselves. As well it includes polynomials such as a(x)(x 2 + 1) b(x)x 3 which generally have higher degree than a(x) and b(x) themselves but which, because of cancellation, could have lower degree. Choose a polynomial m(x) from S with lowest degree. Coming from S, m(x) = a(x)h(x) + b(x)k(x) for some polynomials h(x) and k(x). We now show that m(x) is a common divisor of a(x) and b(x). Let a(x) = m(x)q(x) + r(x) where r(x) = 0 or has lower degree than m(x). Making r(x) the subject of the formula we get: r(x) = a(x) m(x)q(x) = a(x)(1 h(x)q(x)) + b(x)( k(x)q(x)). This means that r(x) belongs to S, but this would contradict the fact that m(x) had lowest degree of any element of S (remember that deg r(x) < deg m(x)). The only way out is for r(x) to be zero, that is, for m(x) to divide a(x). Similarly m(x) must divide b(x) and so it must be a common divisor. But is m(x) a greatest common divisor? Can there be a common divisor having larger degree than that of m(x)? Suppose that d(x) is any common divisor. Since d(x) divides both a(x) and b(x), d(x) must divide m(x) = a(x)h(x) + b(x)k(x). This means that deg d(x) deg m(x). So no common divisor can have larger degree than m(x) and so m(x) is a greatest common divisor. Corollary: All GCDs of a given pair of polynomials are just constant multiples of one another. There is just one monic one among them. 111

12 Proof: Let m(x) = a(x)h(x) + b(x)k(x) be the G.C.D. of a(x) and b(x) obtained as above. Let d(x) be any G.C.D. of a(x) and b(x). Since d(x) divides both a(x) and b(x) it must divide m(x). So deg d(x) deg m(x). But since d(x) is a greatest common divisor, we must have deg d(x) = deg m(x). Now for one polynomial to divide another of the same degree, it must be that they are just constant multiples of one another. Hence all GCDs are constant mulktiples of m(x) and hence of each other. Example 23: The GCD of x 2 1 and x 2 2x + 1 is x 1. The above theorem shows that we can express x 1 in the form (x 2 1)h(x) + (x 2 2x + 1)k(x). Clearly the x 2 terms have to cancel out. A suitable expression is x 1 = (x 2 1)(½) + (x 2 2x + 1)( ½). Here the h(x) and k(x) are constant polynomials, but in more complicated cases they would have higher degree. Theorem 10: For all a, b, q (either integers or polynomials) GCD(a, b) = GCD(a bq, b). Proof: Let d 1 = GCD(a, b) and d 2 = GCD(a bq, b). Then d 1 a bq and d 1 b so d 1 d 2. And since a = (a bq) + bq, d 2 a and d 2 b so d 2 d 1 and d 2 d 1. Since d 1, d 2 are both positive (in the case of integers) or both monic (in the case of polynomials), d 1 = d The Euclidean Algorithm You may recall the Euclidean Algorithm as it applies to finding the GCD of two integers. The same process can be easily adapted to polynomials. It s just that the algebra, when working with polynomials, can get a little messier than the arithmetic, when working with integers. THE EUCLIDEAN ALGORITHM To find GCD(f(x), g(x)): Let c(x) be the remainder on dividing f(x) by g(x). Replace f(x) by g(x). Replace g(x) by c(x). Continue from the top until g(x) 0. When g(x) = 0, make the last non-zero remainder monic. This is the required GCD. Example 24: Find the G.C.D. of the polynomials x 3 1 and x 4 + x 3 x 2 + x 2. Solution: We divide the cubic into the quartic. (Doing it the other way round wouldn t be wrong but all the first division would achieve would be to swap them.) x + 1 x 3 1 ) x 4 + x 3 x 2 + x 2 x 4 x x 3 x 2 + 2x 2 x 3 1 x 2 + 2x 1 112

13 Since we are going to make the answer monic at the end we are entitled to multiply or divide any remainder by any non-zero constant. So we will replace our remainder by x 2 2x + 1. x + 2 x 2 2x + 1 ) x 3 1 x 3 2x 2 + x 2x 2 x 1 2x 2 4x + 2 3x 3 We replace 3x 3 by x 1. x 1 x 1 ) x 2 2x + 1 x 2 x x + 1 x The last non-zero remainder is x 1. This is the required GCD. Example 21 (revisited): The solution to the system of polynomial equations in Example 21 is simply x = 1. Example 25: Find the G.C.D. of the polynomials x and x 4 + x 3 x 2 + x. Solution: x + 1 x ) x 4 + x 3 x 2 + x x 4 + x x 3 x 2 x x 2 1 We change this remainder to x x x ) x x 3 + x x + 1 We change this to x 1. x + 1 x 1 ) x x 2 x x + 1 x

14 The last non-zero remainder (made monic) is 1 so the GCD of these two polynomials is 1. We say that the polynomials have no common factor (which is not strictly true we actually mean that there is no common factor other than constants), or that they are coprime. EXERCISES FOR CHAPTER 9 Exercise 1: Find the degree, the leading coefficient and the roots of the polynomial f(x) = 5x 6 6x 5 x 7. Exercise 2: Which of the following are polynomials in x? (a) x 5 + x; (b) x + x 1 ; (c) 1 + x 2 + x ; (d) 42; (e) x 2 + x + 1. Exercise 3: Find the degree of the polynomial f(x) = (2 n 8)x 2n+1 + 4x 5 + x in terms of n. Exercise 4: Find the quotient and remainder on dividing f(x) = x 4 + 7x + 2 by x 3. Exercise 5: Find the remainder on dividing x 7 + x 4 + x by x 2 3. Exercise 6: Show that x 3 is a factor of the cubic f(x) = x 3 7x x 24. Hence find the roots of f(x). Exercise 7: Given that 1 + i is a root of the polynomial x 4 4x x 2 38x + 34, find its other three roots. Exercise 8: If and are the roots of 4x 2 12x + 7 find the values of: (i) + ; (ii) ; (iii) ( ) 2 ; (iv). Exercise 9: If, and are the roots of the cubic f(x) = x 3 7x x 4, find the value of E = ( + )( + )( + ). Exercise 10: Find the remainder on dividing f(x) = x x 6 5x 2 + x 2 by x 1. Exercise 11: Find the remainder on dividing x 5 2x 3 + 5x 2 7 by x 2 + x + 2. Exercise 12: Given that x = 2 is a root of x 3 6x 2 + 9x 2, find the other two roots. 114

15 Exercise 13: Given that x = i is a root of 6x 4 x 3 + 4x 2 x 2, find the other three roots. Exercise 14: If and are the roots of the quadratic 2x 2 3x + 8, find the values of: (i) ; (ii) ; (iii) Exercise 15: If,, are the roots of x 3 + 5x 2 + 2x 3 find the values of: (i) ; (ii) ; (iii) + +. Exercise 16: If f(x) = x 9 + (k + 1)x 8 + (k 2 + k + 1)x find the sum of the square of the roots of f(x) and hence show that f(x) has at least one none-real root. Exercise 17: Find the GCD of 2x 4 + 7x 3 + 2x 2 + 9x + 7 and 2x 3 + 9x 2 + 9x + 7. Exercise 18: The cubic f(x) = x 3 13x x 80 has a repeated zero. Find it. Exercise 19: Find the GCD of a(x) = 10x 2 + x 21 and b(x) = 8x x Exercise 20: Find the GCD of x and x Exercise 21: Find any multiple roots of the polynomial f(x) = x 4 + 3x 3 + 4x 2 + 3x + 1. Exercise 22: Show that the polynomial f(x) = x 4 2x 3 + 5x 2 4x + 4 is a perfect square (i.e. the square of a polynomial). [HINT: Look for multiple roots.] 115

16 Exercise 1: f(x) = x 7 + 5x 6 6x 5 are 0, 2, 3. SOLUTIONS FOR CHAPTER 9 = x 5 (x 2)(x 3), so deg f(x) = 7; the leading coefficient is 1 and the roots Exercise 2: (a) and (d) are polynomials; the others are not. Exercise 3: For n = 1, f(x) = 6x 3 + 4x 5 + x and so deg f(x) = 5; For n = 2, f(x) = 4x 5 + 4x 5 + x = x so deg f(x) = 2; For n = 3, f(x) = 0x 7 + 4x 5 + x and so deg f(x) = 5; For n 4, deg f(x) = 2n+1. Exercise 4: x 3 + 3x 2 + 9x + 34 x 3 ) x 4 + 7x + 2 x 4 3x 3 3x 3 + 7x + 2 3x 3 9x 2 9x 2 + 7x + 2 9x 2 27x 34x x Hence the quotient is x 3 + 3x 2 + 9x + 34 and the remainder is 104. If we just wanted the reminder it would be much easier to use the Remainder Theorem. The remainder is f(3) = = 104. Exercise 5: x 5 + 3x 3 + x 2 + 9x + 4 x 2 3 ) x 7 + x 4 + x x 7 3x 5 3x 5 + x 4 + x x 5 9x 3 x 4 + 9x 3 + x x 4 3x 2 The remainder is 27x x 3 + 4x x 3 27x 4x x + 1 4x x

17 Exercise 6: We could verify the fact that x 3 is a factor by checking that f(3) = 0. However, since we will need to find the quotient we must carry out the long division process. x 2 4x + 4 x 3 ) x 3 7x x 24 x 3 3x 2 4x x 24 4x x 8x 24 8x 24 0 The quotient is x 2 4x + 8. Solving this quadratic we get x = 2 2i. So the root of f(x) are 3, 2 2i. Exercise 7: Since the polynomial has real coefficients, 1 i the conjugate of 1 + i, must also be a root. Hence x (1 + i) and x (1 i) are factors. Thus (x 1 i)(x 1 + i) = x 2 2x + 2 is a factor. We now divide this into the quartic to obtain the other quadratic factor. x 2 2x + 17 x 2 2x + 2 ) x 4 4x x 2 38x + 34 x 4 2x 3 + 2x 2 2x x 2 38x x 3 + 4x 2 4x 17x 2 34x x 2 34x Hence the polynomial factorises as (x 2 2x + 2)(x 2 2x + 17). Solving x 2 2x + 17 = 0 gives 1 4i. So the other three roots are 1 i, 1 4i. Exercise 8: + = 3, = 7/4. (i) + = = ( + )2 2 = 22 7 ; (ii) = ( + ) = 21/4; (iii) ( ) 2 = ( + ) 2 4 = 9 7 = 2; (iv) = 2. Exercise 9: + + = 7 etc. Now rather than expand E note that + = ( + + ) = 7, and similarly for the others. So E = (7 )(7 )(7 ). Now before you get carried away expanding this out, just note that this f(x) = (x )(x )(x ) and so E is simply f(7) = 73. Exercise 10: The remainder is f(1) =

18 Exercise 11: x 3 x 2 3x + 10 x 2 + x + 2 ) x 5 2x 3 + 5x 2 7 x 5 + x 4 + 2x 3 x 4 4x 3 + 5x 2 7 x 4 x 3 2x 2 Exercise 12: x 2 4x + 1 x 2 ) x 3 6x 2 + 9x 2 x 3 2x 2 4x 2 + 9x 2 4x 2 + 8x x 2 x 2 0 3x 3 + 7x 2 7 3x 3 3x 2 6x 10x 2 6x 7 10x x x 27 Exercise 13: Since i is a root, so is i. Hence (x i)(x + i) = x is a factor. 6x 2 x 2 x ) 6x 4 x 3 + 4x 2 x 2 6x 4 + 6x 2 x 3 2x 2 x 2 x 3 x 2x 2 2 2x Exercise 14: + = 3/2, = 4 (i) = ( + ) = ( + ) = =

19 Exercise 15: + + = 5, + + = 2, = 3. (i) = ( + + ) = 15; (ii) = ( + + ) 2 2( + + ) = 25 4 = 21; (iii) + + = = Exercise 16: = (k + 1) and = k 2 + k + 1. Hence 2 = ( ) 2 2 = (k + 1) 2 2(k 2 + k + 1) = (k 2 + 1). If all the roots were real then 2 0, whereas (k 2 + 1) < 0. Exercise 17: x 1 2x 3 + 9x 2 + 9x + 7 ) 2x 4 + 7x 3 + 2x 2 + 9x + 7 2x 4 + 9x 3 + 9x 2 + 7x 2x 3 7x 2 + 2x + 7 2x 3 9x 2 9x 7 2x x + 14 x 1 2x x + 14 ) 2x 3 + 9x 2 + 9x + 7 2x x x 2x 2 5x + 7 2x 2 11x 14 6x + 21 At the end we will be making the last non-zero remainder monic, so at any stage we can multiply or divide by a non-zero constant to simplify the calculations. So we shall use 2x + 7 instead of 6x x + 2 2x + 7 ) 2x x x 2 + 7x 4x x The last non-zero remainder is 6x + 21 (or equivalently 2x + 7). Making this monic we deduce that the GCD is x + 7/2. 119

20 Exercise 18: A repeated root will be a zero of f (x) = 3x 2 26x + 56 and so will be a zero of GCD(f(x), f (x)). For simplicity let s replace f(x) by 3f(x). x 13/3 3x 2 26x + 56 ) 3x 3 39x x 240 3x 3 26x x 13x x x /3x 728/3 2/3x + 8/3 Replace this remainder by the multiple x 4 = ( 3/2)( 2/3x + 8/3). 3x 14 x 4 ) 3x 2 26x x 2 12x 14x x Exercise 19: Rather than divide we can first subtract. This will simplify the arithmetic. GCD(a(x), b(x)) = GCD(a(x) b(x), b(x)) = GCD(2x 2 29x 48, 8x x + 27). = GCD(2x 2 29x 48, (8x x + 27) 4(2x 2 29x 48)) = GCD(2x 2 29x 48, 146x + 219) = GCD(2x 2 29x 48, 2x + 3) since 146x = 73(2x + 3). x 16 2x + 3 ) 2x 2 29x 48 2x 2 + 3x 32x 48 32x 48 0 Exercise 20: x x ) x x 5 + x x + 1 Replace this remainder by x

21 x 3 + x 2 + x x 1 ) x x 4 x 3 x x 3 x 2 x x Making this monic we see that the GCD is 1. Exercise 21: f (x) = 4x 3 + 9x 2 + 8x + 3. For convenience replace f(x) by by 4f(x). x + 3/4 4x 3 + 9x 2 + 8x + 3 ) 4x x x x + 4 4x 4 + 9x 3 + 8x 2 + 3x 3x 3 + 8x 2 + 9x + 4 3x 3 + (27/4)x 2 + 6x + 9/4 (5/4)x 2 + 3x + 7/4 (4/5)x 3/25 5x x + 7 ) 4x 3 + 9x 2 + 8x + 3 4x 3 + (48/5)x 2 + (28/5)x ( 3/5)x 2 + (12/5)x + 3 ( 3/5)x 2 (36/25)x 21/25 (96/25)x + 96/25 5x + 7 x + 1 ) 5x x + 7 5x 2 + 5x 7x + 7 7x Exercise 22: f (x) = 4x 3 6x x x x 3 3x 2 + 5x 2) x 4 2x 3 + 5x 2 4x + 4 x 4 1.5x x 2 x 0.5x x 2 3x x x x x x Replace this by x 2 x

22 2x 1 x 2 x + 2 ) 2x 3 3x 2 + 5x 2 2x 3 2x 2 + 4x x 2 + x 2 x 2 + x

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