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11.Remainder and Factor Theorem

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11: THE REMAINDER AND FACTOR THEOREM Solving and simplifying polynomials Thus, from arithmetic, we know that we can express a In our study of quadratics, one of the methods used to dividend as: simplify and solve was factorisation. For example, we may solve for x in the following equation as follows: Dividend = (Quotient ´ Divisor) + Remainder 2 x + 5x + 6 = 0 When there is no remainder, � = 0, and the divisor is (x + 3)(x + 2) = 0 now a factor of the number, so x + 3 = 0,⇒ x = −3 x + 2 = 0 ⇒ x = −2 Dividend = Quotient ´ Factor Hence, x = −3 or −2 are solutions or roots of the The process we followed in arithmetic when dividing quadratic equation. is very similar to what is to be done in algebra. Examine the following division problems in algebra A more general name for a quadratic is a polynomial and note the similarities. of degree 2, since the highest power of the unknown is two. The method of factorisation worked for Division of a polynomial by a linear expression quadratics whose solutions are integers or rational We can apply the same principles in arithmetic to numbers. dividing algebraic expressions. Let the quadratic For a polynomial of order 3, such as function �(�) represent the dividend, and (� − 1) the x3 + 4x2 + x − 6 = 0 the method of factorisation may divisor, where - also be applied. However, obtaining the factors is not �(�) = 3� − � + 2 as simple as it was for quadratics. We would likely have to write down three linear factors, which may Quotient 3x 2 + prove difficult. In this section, we will learn to use the 2 Dividend remainder and factor theorems to factorise and to solve x −1 3x − x + 2 polynomials that are of degree higher than 2. − (3x2 − 3x) Divisor 2x 2 + Before doing so, let us review the meaning of basic − (2x − 2) terms in division. 4 Remainder Terms in division We are familiar with division in arithmetic. From the above example, we can deduce that: 3�- − � + 2 = (3� + 2)(� − 1) + 4 Quotient 15 ↑ ↑ ↑ ↑ Dividend 32 489 Dividend Quotient Divisor Remainder 32 Divisor 169 Consider (� − 1) = 0, 160 �(�) = (3� + 2) × 0 + 4 = 4 9 Remainder But, (� − 1) = 0 implies that � = 1 The number that is to be divided is called the dividend. The dividend is divided by the divisor. Therefore, when � = 1, �(�) = 4 or �(1) = 4. The result is the quotient and the remainder is what is left over. We can conclude that when the polynomial 3�- − � + 2 From the above example, we can deduce that: is divided by (� − 1), the remainder is 489 = (15 × 32) + 9 �(1) = 4 ↑ ↑ ↑ ↑ Dividend Quotient Divisor Remainder We shall now perform division using a cubic polynomial as our dividend. 2 We can conclude that when the polynomial x +14x + 25 4 - 3 2 2� − 3� + 4� + 5 x − 2 x +12x − 3x + 4 is divided by (� + 2), the remainder is − (x3 − 2x2 ) �(−2) = −31 14x2 − 3x + 4 − (14x − 28x) The Remainder Theorem for divisor (� − �) 25x + 4 From the above examples, we saw that a polynomial −(25x − 50) can be expressed as a product of the quotient and the divisor plus the remainder: 54 3�- − � + 2 = (3� + 2)(� − 1) + 4 From the above example, we can deduce that: �4 + 12�- − 3� + 4 = (�- + 14� + 25)(� − 2) + 54 4 - - � + 12� − 3� + 4 = (� + 14� + 25)(� − 2) + 54 ↑ ↑ ↑ ↑ 2�4 − 3�- + 4� + 5 = (2�- + � + 18)(� + 2) − 31 Dividend Quotient Divisor Remainder We can now formulate the following expression Consider (� − 2) = 0, f (x) Q(x) �(�) = (�- + 14� + 25) × 0 + 54 = 54 where, is a polynomial whose quotient is and whose remainder is R when divided by (x − a) . But, (� − 2) = 0 implies that � = 2 fx()= Qx ()´- ( x a )+ R Therefore, when � = 2, �(�) = 54 or �(2) = 54. If we were to substitute � = � in the above expression, We can conclude that when the polynomial then our result will be equal to �, the remainder when �4 + 12�- − 3� + 4 the divisor, (� − �) is divided by the polynomial, is divided by (� − 2), the remainder is �(�). �(2) = 54 We are now able to state the remainder theorem. Now consider another example of a cubic polynomial divided by a linear divisor. The Remainder Theorem If f (x) is any polynomial and f (x) is divided 2 by , then the remainder is 2x + 7x +18 (x − a) fa(). x + 2 2x3 − 3x2 + 4x + 5 The validity of this theorem can be tested in any of − (2x3 + 4x2 ) the equations above, for example: − 7x2 + 4x + 5 - − (7x2 −14x) 1. When 3� − � + 2 was divided by (� − 1), the remainder was 4. 18x + 5 −(18x + 36) According to the remainder theorem, the remainder − 31 can be computed by substituting � = 1 in �(�) �(�) = 3�- − � + 2 �(1) = 3(1)- − 1 + 2 From the above example, we can deduce that: �(1) = 4 4 - - 2� − 3� + 4� + 5 = (2� + � + 18)(� + 2) − 31 ↑ ↑ ↑ ↑ 2. When �4 + 12�- − 3� + 4 was divided by Dividend Quotient Divisor Remainder (� − 2), the remainder was 54. Consider (� + 2) = 0, According to the remainder theorem, the remainder �(�) = (2�- + � + 18) × 0 − 31 = −31 can be computed by substituting � = 2 in �(�) �(�) = �4 + 12�- − 3� + 4 But, (� + 2) = 0 implies that � = −2 �(2) = (2)4 + 12(2)- − 3(2) + 4 Therefore, when � = −2, �(�) = −31 or �(2) = 8 + 48 − 6 + 4 �(−2) = −31. �(2) = 54 The above rule is called the Factor Theorem, it is a 3. When 2�4 − 3�- + 4� + 5 was divided by special case of the Remainder Theorem, when � = 0. (� + 2) the remainder was −31. The validity of this theorem can be tested by substituting � = 1 in each of the above functions. According to the remainder theorem, the remainder can be computed by substituting � = −2 in �(�) �(�) = 3�- − � − 2 �(�) = 8�4 − 10�- − � + 3 - 4 - �(�) = 2�4 − 3�- + 4� + 5 �(1) = 3(1) − 1 − 2 �(1) = 8(1) − 10(1) − 1 4 - + 3 �(−2) = 2(−2) − 3(−2) + 4(−2) + 5 �(1) = 0 �(−2) = −16 − 12 − 8 + 5 �(1) = 0 �(−2) = −31 Example 1 In the above examples, when we let (� − 1) = 0, or Find the remainder when � = 1, �(�) = 0 because the remainder, � = 0. �(�) = 3�4 + �- − 4� − 1 is divided by (� − 2). The Factor Theorem If is any polynomial and is divided Solution f (x) f (x) by (x − a) , and the remainder fa()= 0 then By the Remainder Theorem, the remainder is �(2). (x − a) is a factor of f (x) �(�) = 3�4 + �- − 4� − 1 �(2) = 3(2)4 + (2)- − 4(2) − 1 �(2) = 24 + 4 − 8 − 1 Example 2 �(2) = 19 Show that (� − 2) is a factor of Hence, the remainder is 19 4 - �(�) = 3� + � − 14� The Factor Theorem for divisor (� − �) Solution Now, consider the following examples when there is no remainder. By the Factor Theorem, if (� − 2) is a factor of the remainder is zero. We now compute the 2 remainder, �(2). 3x + 2 8x − 2x − 3 �(�) = 3�4 + �- − 14� 2 x −1 3x − x − 2 x −1 8x3 −10x2 − x + 3 �(2) = 3(2)4 + (2)- − 14(2) 2 3 2 �(2) = 24 + 4 − 28 − (3x − 3x) (8x 8x ) − − �(2) = 0 2x − 2 2 − 2x − x + 3 (2x 2) 2 − − − (−2x + 2x) Hence, (� − 2) is a factor of �(�). 0 − 3x + 3 −(3x + 3) 0 The Remainder and Factor Theorem for divisor (�� + �) (x a) When the divisor is not in the form, − , but in the We can express the dividend as a product of the divisor general linear form (�� + �), the remainder can no and the quotient only, since � = 0. longer be �(�). This is because the coefficient of x is not equal to one. - �(�) = 3� − � − 2 = (3� + 2)(� − 1) �(�) = 8�4 − 10�- − � + 3 = (8�- − 2� − 3)(� − 1) Consider the following example, where the divisor is of the form, (�� + �). We can now formulate the following expression where �(�) is a polynomial, �(�) is the quotient and (� − �) Let (2� + 3) be a divisor of is a factor of the polynomial. �(�) = 2�4 + 7�- + 2� + 9 We perform the division as shown below and note �(�) = �(�) × (� − �) that the remainder is 15. 2 x + 2x − 2 where, f (x) is a polynomial whose quotient is Q(x) 2x + 3 2x3 + 7x2 + 2x + 9 and the remainder is R when divided by (�� + �). 3 2 By setting (�� + �) = 0, the above polynomial will − (2x + 3x ) 2 become 4x + 2x + 9 ( ) 2 � � = � D − (4x + 6x) When (�� + �) = 0, � = − . − 4x + 9 E Now, since −(4x − 6) D �(�) = �, when � = − , we conclude that 15 E � � F− G = � � We can deduce that: We are now in a position to restate the remainder theorem when the divisor is of the form ax+ b . 2�4 + 7�- + 2� + 9 = (�- + 2� − 2)(2� + 3) + 15 ( ) Consider (2� + 3) = 0, The Remainder Theorem �(�) = (�- + 2� − 2) × 0 + 15 = 15 If fx( ) is any polynomial and fx( ) is divided by 4 But, (2� + 3) = 0 implies that � = − . - æöb 4 ax+ b then the remainder is f - . Therefore, when � = − , �(�) = 15 or ( ) ç÷ - èøa 4 � A− B = 15. æöb - If f ç÷- = 0, then (ax+ b)is a factor of fx( ) . We can conclude that when the polynomial èøa 4 - 2� + 7� + 2� + 9 is divided by (2� + 3), the 4 remainder is � A− B = 15 We apply the Remainder Theorem to obtain the - 4 - Now consider the example below. remainder when �(�) = 2� + 7� + 2� + 9 was divided by (2� + 3). 2 By the Remainder Theorem, the remainder is 3x + 6x −1 4 3 2 � A− B. 3x −1 9x +15x − 9x +1 - 3 3 3 3 3 2 � F− G = 2(− )4 + 7(− )- + 2(− ) + 9 − (9x − 3x ) 2 2 2 2 2 3 27 63 18x − 9x +1 � F− G = − + − 3 + 9 2 2 4 4 − (18x − 6x) 3 3x 1 � F− G = 15 − + 2 − (−3x +1) We can apply the Factor Theorem to show that 0 (3� − 1) is a factor of �(�), where �(�) = 9�4 + 15�- − 9� + 1 We can deduce that: By the Factor Theorem (3� − 1) is a factor of �(�).
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    Lyashko-Looijenga Morphisms and Submaximal Factorisations of A

    LYASHKO-LOOIJENGA MORPHISMS AND SUBMAXIMAL FACTORIZATIONS OF A COXETER ELEMENT VIVIEN RIPOLL Abstract. When W is a finite reflection group, the noncrossing partition lattice NC(W ) of type W is a rich combinatorial object, extending the notion of noncrossing partitions of an n-gon. A formula (for which the only known proofs are case-by-case) expresses the number of multichains of a given length in NC(W ) as a generalized Fuß-Catalan number, depending on the invariant degrees of W . We describe how to understand some specifications of this formula in a case-free way, using an interpretation of the chains of NC(W ) as fibers of a Lyashko-Looijenga covering (LL), constructed from the geometry of the discriminant hypersurface of W . We study algebraically the map LL, describing the factorizations of its discriminant and its Jacobian. As byproducts, we generalize a formula stated by K. Saito for real reflection groups, and we deduce new enumeration formulas for certain factorizations of a Coxeter element of W . Introduction Complex reflection groups are a natural generalization of finite real reflection groups (that is, finite Coxeter groups realized in their geometric representation). In this article, we consider a well-generated complex reflection group W ; the precise definitions will be given in Sec. 1.1. The noncrossing partition lattice of type W , denoted NC(W ), is a particular subset of W , endowed with a partial order 4 called the absolute order (see definition below). When W is a Coxeter group of type A, NC(W ) is isomorphic to the poset of noncrossing partitions of a set, studied by Kreweras [Kre72].