# 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.

 To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video
Save this PDF as:

Size: px
Start display at page:

Download "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."

## Transcription

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

2 9.2. Degree of a Polynomial The degree of a polynomial is the largest power of x that occurs with a non-zero coefficient. That is, if a(x)=a n x n + a n 1 x n a 1 x + a 0, the degree of a(x) is n (provided that a n 0). We can write deg a(x) = n. Examples 4: Polynomials of degree 2 are quadratics, of the form ax 3 + bx+c (where a 0). Polynomials of degree 1 are the so-called linear polynomials such as 2x + 3 and ½x ¼. Polynomials of degree 0 are the non-zero constant polynomials. It might seem strange for numbers such as 3 or ½ to be considered as polynomials, but they can. If it makes you feel better you can write 3 as 0x 2 + 0x + 3. (Remember that perhaps you once found it strange to call 3 a fraction until you learnt that it could be written as 3 1.) There is one polynomial for which the degree remains undefined. It is the zero constant polynomial, 0. (Some books define this degree to be for certain technical reasons but we shall leave it undefined.) The coefficient of x n, for a polynomial of degree n, is called its leading coefficient. For example the leading coefficient of 3x 2 x + 5 is 3. But beware. The leading coefficient does not always come first the leading coefficient of 1 x 2 is 1, not 1. A monic polynomial is one where the leading coefficient is 1. Clearly every non-zero polynomial can be made monic by dividing it by its leading coefficient. Example 5: The polynomial 4x 3 8x +1 has degree 3. Its leading coefficient is 4 and so it is not monic. However it can be expressed as 4 times the monic polynomial x 3 2x + ¼. If F is a field (for example F might be Q, the system of rational numbers) we denote the set of all polynomials with coefficients coming from F by the symbol F[x]. Examples 6: For example R[x] contains the polynomial x 2 2x + 3 but not x 2 + i. However x 2 + i is in C[x] Addition and Multiplication of Polynomials Polynomials are added (and subtracted) in the usual way just add (or subtract) the corresponding coefficients. Multiplication is somewhat more complicated to describe abstractly, yet it just amounts to expanding the product of the expressions in the way you have always done. For example (ax 2 + bx + c)(dx + e) = adx 3 + (ae + bd)x 2 + (be + cd)x + ce. With these operations the system F[x] behaves very much like a field itself, but with one very important difference. In a field nearly every number has an inverse under multiplication (in fact 0 is the only exception). Most polynomials, on the other hand, do not have inverses. For example, since 1 x and 1 1 x are not polynomials, the polynomials x and 1 x do not have (polynomial) inverses. In fact the only polynomials with inverses are the non-zero constant polynomials such as 2 (whose inverse is the constant polynomial ½). 102

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

8 But from this formula we can compute the roots of a quadratic over any field (except those, such as Z 2 where 1+1=0). And since it is possible, and not very difficult, to find the square roots of any complex number we can find roots for any complex quadratic. Example 18: Find the zeros of the quadratic x 2 + 2ix + i. Solution: x = 2i 4 4i 2 and since 4 4i = 32(cos 5 i/4 + i sin 5 i/4) we can express its square roots as 32 ¼ (cos 5 i/8 + i sin 5 i/8). The roots of the quadratic are thus 32 ¼ cos 5 i/8 + (32 ¼ sin 5 i/8 1)i. The Fundamental Theorem states that not only complex quadratics but complex cubics, complex quartics,... have complex roots. Because of this we say that C is algebraically closed. No new numbers need to be invented. The history of number consisted of mathematicians continuing to extend the number system so as to provide roots for polynomials which would otherwise have none. God, it has been said, gave us the natural numbers 1,2,3,... and man created the rest. When we just had the natural numbers 1,2,3,... we had to say that the equation 2x = 1 has no solutions. So fractions were invented. Then 2x = 1 could be solved but x 2 = 2 could not. So along come the irrational numbers. Since 2 now exists, x 2 = 2 now has a solution (just one because so far there is no such thing as a negative number). This all happened back in classical times. But despite all these extra numbers that had been invented, an equation such as x + 2 = 1 still had no solution. Negative numbers took another thousand years or so arrive. Once they did then x + 2 = 1 was OK. But x = 0 could not be solved. There were no roots for x The last stage was the extension from the real numbers to the complex numbers which took place in the 17th century. Now a polynomial equation such as x = 0 could be solved, but what about x 2 + i = 0, or x 5 +ix 3 = 1 + 2i? Might we not have to go on inventing more and more numbers? Might not our number system spring new leaks whenever we try to patch up existing ones? After all, whenever we extend our number system we introduce new equations. But the marvellous thing is that having arrived at the system of complex numbers we had, in a sense, reached perfection. No new numbers are necessary. We can solve all polynomial equations, with complex coefficients, entirely from within the complex number system itself. There is a complex number x such that x 2 + i = 0 (in fact two of them). There is a complex number x such that x 5 +ix 3 = 1 + 2i (in fact five such solutions). The Fundamental Theorem shows that we need not extend our number system any further. As a result of the Fundamental Theorem of Algebra every polynomial with complex coefficients can be factorised completely into linear factors. If the coefficients are real, and we insist that the factors have real coefficients, we cannot expect to be able to factorise the polynomial quite so completely. However we can factorise any real polynomial into real factors which, at worst, are quadratic. This comes about because the non-real roots of a real polynomial split into conjugate pairs. 108

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

### SOLVING POLYNOMIAL EQUATIONS

C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)

ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson

JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials

### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

### 3.6 The Real Zeros of a Polynomial Function

SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix,

### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.

Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Factoring Polynomials

Factoring Polynomials Hoste, Miller, Murieka September 12, 2011 1 Factoring In the previous section, we discussed how to determine the product of two or more terms. Consider, for instance, the equations

### 3.4 Complex Zeros and the Fundamental Theorem of Algebra

86 Polynomial Functions.4 Complex Zeros and the Fundamental Theorem of Algebra In Section., we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and

### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### Application. Outline. 3-1 Polynomial Functions 3-2 Finding Rational Zeros of. Polynomial. 3-3 Approximating Real Zeros of.

Polynomial and Rational Functions Outline 3-1 Polynomial Functions 3-2 Finding Rational Zeros of Polynomials 3-3 Approximating Real Zeros of Polynomials 3-4 Rational Functions Chapter 3 Group Activity:

### 3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes

Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### Functions and Equations

Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c

### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

### H/wk 13, Solutions to selected problems

H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.

### March 29, 2011. 171S4.4 Theorems about Zeros of Polynomial Functions

MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 4: Polynomial and Rational Functions 4.1 Polynomial Functions and Models 4.2 Graphing Polynomial Functions 4.3 Polynomial

### Introduction to Finite Fields (cont.)

Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number

### UNCORRECTED PAGE PROOFS

number and and algebra TopIC 17 Polynomials 17.1 Overview Why learn this? Just as number is learned in stages, so too are graphs. You have been building your knowledge of graphs and functions over time.

### FACTORISATION YEARS. A guide for teachers - Years 9 10 June 2011. The Improving Mathematics Education in Schools (TIMES) Project

9 10 YEARS The Improving Mathematics Education in Schools (TIMES) Project FACTORISATION NUMBER AND ALGEBRA Module 33 A guide for teachers - Years 9 10 June 2011 Factorisation (Number and Algebra : Module

### Cubic Functions: Global Analysis

Chapter 14 Cubic Functions: Global Analysis The Essential Question, 231 Concavity-sign, 232 Slope-sign, 234 Extremum, 235 Height-sign, 236 0-Concavity Location, 237 0-Slope Location, 239 Extremum Location,

### 2.3. Finding polynomial functions. An Introduction:

2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned

### Basics of Polynomial Theory

3 Basics of Polynomial Theory 3.1 Polynomial Equations In geodesy and geoinformatics, most observations are related to unknowns parameters through equations of algebraic (polynomial) type. In cases where

### Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.

Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### Algebra I Vocabulary Cards

Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression

### RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### Copyrighted Material. Chapter 1 DEGREE OF A CURVE

Chapter 1 DEGREE OF A CURVE Road Map The idea of degree is a fundamental concept, which will take us several chapters to explore in depth. We begin by explaining what an algebraic curve is, and offer two

### 3.6. The factor theorem

3.6. The factor theorem Example 1. At the right we have drawn the graph of the polynomial y = x 4 9x 3 + 8x 36x + 16. Your problem is to write the polynomial in factored form. Does the geometry of the

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### Zeros of Polynomial Functions

Review: Synthetic Division Find (x 2-5x - 5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 3-5x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 3-5x 2 + x + 2. Zeros of Polynomial Functions Introduction

### COLLEGE ALGEBRA. Paul Dawkins

COLLEGE ALGEBRA Paul Dawkins Table of Contents Preface... iii Outline... iv Preliminaries... Introduction... Integer Exponents... Rational Exponents... 9 Real Exponents...5 Radicals...6 Polynomials...5

### 3.2 The Factor Theorem and The Remainder Theorem

3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial

### Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

### MATH 65 NOTEBOOK CERTIFICATIONS

MATH 65 NOTEBOOK CERTIFICATIONS Review Material from Math 60 2.5 4.3 4.4a Chapter #8: Systems of Linear Equations 8.1 8.2 8.3 Chapter #5: Exponents and Polynomials 5.1 5.2a 5.2b 5.3 5.4 5.5 5.6a 5.7a 1

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### Breaking The Code. Ryan Lowe. Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and

Breaking The Code Ryan Lowe Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and a minor in Applied Physics. As a sophomore, he took an independent study

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### An Introduction to Galois Fields and Reed-Solomon Coding

An Introduction to Galois Fields and Reed-Solomon Coding James Westall James Martin School of Computing Clemson University Clemson, SC 29634-1906 October 4, 2010 1 Fields A field is a set of elements on

### Polynomial Expressions and Equations

Polynomial Expressions and Equations This is a really close-up picture of rain. Really. The picture represents falling water broken down into molecules, each with two hydrogen atoms connected to one oxygen

### Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called

### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com

Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).

### Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

### Zeros of a Polynomial Function

Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we

### 5.4 The Quadratic Formula

Section 5.4 The Quadratic Formula 481 5.4 The Quadratic Formula Consider the general quadratic function f(x) = ax + bx + c. In the previous section, we learned that we can find the zeros of this function

### The Method of Partial Fractions Math 121 Calculus II Spring 2015

Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for

### Lecture Notes on Polynomials

Lecture Notes on Polynomials Arne Jensen Department of Mathematical Sciences Aalborg University c 008 Introduction These lecture notes give a very short introduction to polynomials with real and complex

### The Notebook Series. The solution of cubic and quartic equations. R.S. Johnson. Professor of Applied Mathematics

The Notebook Series The solution of cubic and quartic equations by R.S. Johnson Professor of Applied Mathematics School of Mathematics & Statistics University of Newcastle upon Tyne R.S.Johnson 006 CONTENTS

### 1 Lecture: Integration of rational functions by decomposition

Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.

### Solving Cubic Polynomials

Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial

### calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,

Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials

### Rational Polynomial Functions

Rational Polynomial Functions Rational Polynomial Functions and Their Domains Today we discuss rational polynomial functions. A function f(x) is a rational polynomial function if it is the quotient of

### CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation

CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of

### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style

Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with

### Understanding Basic Calculus

Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other

### expression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method.

A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### z = i ± 9 2 2 so z = 2i or z = i are the solutions. (c) z 4 + 2z 2 + 4 = 0. By the quadratic formula,

91 Homework 8 solutions Exercises.: 18. Show that Z[i] is an integral domain, describe its field of fractions and find the units. There are two ways to show it is an integral domain. The first is to observe:

### SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS

SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS Assume f ( x) is a nonconstant polynomial with real coefficients written in standard form. PART A: TECHNIQUES WE HAVE ALREADY SEEN Refer to: Notes 1.31

### Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

### 5. Factoring by the QF method

5. Factoring by the QF method 5.0 Preliminaries 5.1 The QF view of factorability 5.2 Illustration of the QF view of factorability 5.3 The QF approach to factorization 5.4 Alternative factorization by the

### Algebra 1 Course Title

Algebra 1 Course Title Course- wide 1. What patterns and methods are being used? Course- wide 1. Students will be adept at solving and graphing linear and quadratic equations 2. Students will be adept

### Introduction to Diophantine Equations

Introduction to Diophantine Equations Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles September, 2006 Abstract In this article we will only touch on a few tiny parts of the field

### 1.3 Algebraic Expressions

1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,

### CONTINUED FRACTIONS AND FACTORING. Niels Lauritzen

CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents

### Vocabulary Words and Definitions for Algebra

Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms

### CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA

We Can Early Learning Curriculum PreK Grades 8 12 INSIDE ALGEBRA, GRADES 8 12 CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA April 2016 www.voyagersopris.com Mathematical

### Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem)

Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem) In order to understand the details of the Fingerprinting Theorem on fingerprints of different texts from Chapter 19 of the

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### Polynomials and Factoring

Lesson 2 Polynomials and Factoring A polynomial function is a power function or the sum of two or more power functions, each of which has a nonnegative integer power. Because polynomial functions are built

### 1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).

.7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational

### 3. INNER PRODUCT SPACES

. INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

### a = bq + r where 0 r < b.

Lecture 5: Euclid s algorithm Introduction The fundamental arithmetic operations are addition, subtraction, multiplication and division. But there is a fifth operation which I would argue is just as fundamental

### It is time to prove some theorems. There are various strategies for doing

CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it

### POLYNOMIAL FUNCTIONS

POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a

### Decomposing Rational Functions into Partial Fractions:

Prof. Keely's Math Online Lessons University of Phoenix Online & Clark College, Vancouver WA Copyright 2003 Sally J. Keely. All Rights Reserved. COLLEGE ALGEBRA Hi! Today's topic is highly structured and

### 5 Indefinite integral

5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse

### 3 Factorisation into irreducibles

3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could

### 5.1 The Remainder and Factor Theorems; Synthetic Division

5.1 The Remainder and Factor Theorems; Synthetic Division In this section you will learn to: understand the definition of a zero of a polynomial function use long and synthetic division to divide polynomials

### Zeros of Polynomial Functions

Zeros of Polynomial Functions Objectives: 1.Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions 2.Find rational zeros of polynomial functions 3.Find conjugate

### Mth 95 Module 2 Spring 2014

Mth 95 Module Spring 014 Section 5.3 Polynomials and Polynomial Functions Vocabulary of Polynomials A term is a number, a variable, or a product of numbers and variables raised to powers. Terms in an expression

### FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.

FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is