Polynomials. In many problems, it is useful to write polynomials as products. For example, when solving equations: Example:

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1 Polynomials Monomials: 10, 5x, 3x 2, x 3, 4x 2 y 6, or 5xyz 2. A monomial is a product of quantities some of which are unknown. Polynomials: x 3x 2 + x 3, or 4x 2 y 6 + 5xyz 2. A polynomial is a sum of one or more monomials. Mono = one, Poly = many. 1. Factorizations In many problems, it is useful to write polynomials as products. For example, when solving equations: Example: A rectangular house of area 1000m 2 has been enlarged with a corridor or widths 15 m 2 and 10 m 2 like in the picture, resulting in a total area of 1925 m 2. What were the initial width and length of the house? 15 x y 10 (x + 15)(y + 10) = 1925 and xy = 1000 xy + 15y + 10x = 1925 and xy = y + 10x = 775 and xy = x = y and (775 15y)y = 10, 000 x = y and 0 = 15y 2 775y + 10, 000 hence 0 = 3y 2 155y Search (3y )(y ) hence 0 = (3y 80)(y 25) Hence y = 25 and x = = 40 or y = 80 3 and x = = Note: We were lucky to be able to factor 3y 2 155y+2000 = (3y )(y ). We looked among the factors = 2000 at a pair small enough such that 3 + = 155. The pair is too large while is just right. Practice Factorisation by Grouping: a) Solve xy 5x + 7y 35 = 0. 1

2 2 b) Find all integers x, y such that 2xy 3x 18y = 1. c) Factorise x 5 4x 3 y 2 2x 2 + 8y 2. d) Factorise 3x 2 + 2x 8. Solutions: a) xy 5x + 7y 35 = x(y 5) + 7(y 5) = (x + 7)(y 5). Hence x = 7 or y = 5. b) 1 = 2xy 3x 18y = x(2y 3) 9 2y. We need to complete the second term to 9(2y 3) by adding 27. Hence 28 = 2xy 3x 18y + 27 = x(2y 3) 9(2y 3) = (x 9)(2y 3). Keeping in mind that 2y 3 is odd, we can only have 2y 3 = 1, x 9 = 28 hence y = 2, x = 37. 2y 3 = 1, x 9 = 28 hence y = 1, x = 19. 2y 3 = 7, x 9 = 4 hence y = 5, x = 13. 2y 3 = 7, x 9 = 4 hence y = 2, x = 5. c) x 5 4x 3 y 2 2x 2 + 8y 2 = x 3 (x 2 4y 2 ) 2(x 2 4y 2 ) = (x 3 2)(x 2 4y 2 ) = (x 3 2)(x 2y)(x + 2y). d) 3x 2 + 2x 8 = (3x + )(x + ) where = 8 so we may try 2 ( 4), 2 4, 1 ( 8), 1 8. We get 3x 2 + 2x 8 = (3x 4)(x + 2). Grouping and Rearrangement Formulae: Suppose we have two strings of numbers a 1 a 2... a n and b 1 b 2... b n. Among all the sums of products a i b j, where each a is paired with a different b, we want to find the largest sum. a) If a 1 a 2 and b 1 b 2, prove that a 1 b 1 + a 2 b 2 a 1 b 2 + a 2 b 1. b) If a 1 a 2 a 3 and b 1 b 2 b 3, prove that a 1 b 1 + a 2 b 2 + a 3 b 3 a 1 b 2 + a 2 b 3 + a 3 b 1 a 1 b 3 + a 2 b 2 + a 1 b 3, a 1 b 1 + a 2 b 2 + a 3 b 3 a 1 b 3 + a 2 b 1 + a 3 b 2 a 1 b 3 + a 2 b 2 + a 1 b 3. c) In general, if a 1 a 2... a n and b 1 b 2... b n, prove that a 1 b 1 + a 2 b 2 + a 3 b a n b n is the largest among the sums and a 1 b n + a 2 b n a n 1 b 2 + a n b 1 is the smallest among the sums. Solution: a) To compare a 1 b 1 +a 2 b 2 and a 1 b 2 +a 2 b 1, we check the difference: as both factors are 0. a 1 b 1 + a 2 b 2 a 1 b 2 a 2 b 1 = a 1 b 1 a 1 b 2 + a 2 b 2 a 2 b 1 = a 1 (b 1 b 2 ) a 2 (b 1 b 2 ) = (a 1 a 2 )(b 1 b 2 ) 0

3 3 b) Similarly a 1 b 1 + a 2 b 2 + a 3 b 3 a 1 b 2 a 2 b 3 a 3 b 1 = a 1 b 1 a 1 b 2 + a 2 b 2 a 2 b 3 + a 3 b 3 a 3 b 1 = a 1 (b 1 b 2 ) + a 2 (b 2 b 3 ) + a 3 (b 3 b 1 ). Note that (b 1 b 2 ) 0 and (b 2 b 3 ) 0 but (b 3 b 1 ) 0. Moreover, (b 3 b 1 ) = (b 1 b 3 ) = (b 1 b 2 + b 2 b 3 ) = (b 1 b 2 ) (b 2 b 3 ). Hence by grouping again: a 1 b 1 + a 2 b 2 + a 3 b 3 a 1 b 2 a 2 b 3 a 3 b 1 = a 1 (b 1 b 2 ) + a 2 (b 2 b 3 ) + a 3 (b 3 b 1 ) = a 1 (b 1 b 2 ) + a 2 (b 2 b 3 ) a 3 (b 1 b 2 ) a 3 (b 2 b 3 ) = (a 1 a 3 )(b 1 b 2 ) + (a 2 a 3 )(b 2 b 3 ) 0. Similary with the other group. c) Consider a random sum and assume that a 1 b 1 is not in it. Then there are some terms a 1 b j and a k b 1 in it. Then (a 1 b 1 + a k b j ) (a 1 b j + a k b 1 ) = (a 1 a k )(b 1 b j ) 0 so we may increase the sum by replacing a 1 b j + a k b 1 with a 1 b 1 + a k b j. Now we only have (n 1) pairs of numbers which might not be optimally matched, but by applying the previous step for the other pairs a i b i one at a time we obtain the largest sum a 1 b 1 + a 2 b 2 + a 3 b a n b n. The smallest case works out similarly. 2. Factorization Formulae Warm-up: x 2 y 2 = (x y)(x + y) and x 2 + 2xy + y 2 = (x + y) 2. More formulae: Check the formulae below by multiplying through the right-hand-side, using distributivity: n; x n y n = (x y)(x n 1 + x n 2 y xy n 2 + y n 1 ) for positive integers x n + y n = (x y)(x n 1 x n 2 y +... xy n 2 + y n 1 ) for n odd. What can you say about the right-hand-side when n is even? For example: x 2 y 2 = (x y)(x + y); x 3 y 3 = (x y)(x 2 + xy + y 2 ); x 4 y 4 = (x y)(x 3 + x 2 y + xy 2 + y 3 ); x 5 y 5 = (x y)(x 4 + x 3 y + x 2 y 2 + xy 3 + y 4 ); and x 3 + y 3 = (x + y)(x 2 xy + y 2 ); x 5 + y 5 = (x + y)(x 4 x 3 y + x 2 y 2 xy 3 + y 4 ). Example: Factoring polynomials by combining factorization formulae: x 4 y 4 = (x 2 y 2 )(x 2 + y 2 ) = (x y)(x + y)(x 2 + y 2 );

4 4 x 4 + y 4 + x 2 y 2 = x 4 + y 4 + 2x 2 y 2 x 2 y 2 = (x 2 + y 2 ) 2 x 2 y 2 = (x 2 + y 2 xy)(x 2 + y 2 + xy); x 4 + y 4 = x 4 + y 4 + 2x 2 y 2 2x 2 y 2 = (x 2 + y 2 ) 2 2x 2 y 2 = (x 2 + y 2 2xy)(x 2 + y 2 + 2xy); x 6 y 6 = (x 3 y 3 )(x 3 + y 3 ) = (x y)(x 2 + xy + y 2 )(x + y)(x 2 xy + y 2 ). x 7 + x = (x 7 x) + x + x = x(x 6 1) + x 2 + x + 1 = x(x 3 1)(x 3 + 1) + x 2 + x + 1 = x(x 1)(x 2 + x + 1)(x 3 + 1) + x 2 + x + 1 = (x 2 + x + 1)[x(x 1)(x 3 + 1) + 1] The binomial formula: x n + C n 1 x n 1 y + C n 2 x n 2 y C n n 2x 2 y n 2 + C n n 1xy n 1 + y n = (x + y) For example: x 2 + 2xy + y 2 = (x + y) 2 ; x 3 + 3x 2 y + 3xy 2 + y 3 = (x + y) 3 ; x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 = (x + y) 4 ; x 5 + 5x 4 y + 10x 3 y x 2 y 3 + 5xy 4 + y 5 = (x + y) 5 ; x 2 2xy + y 2 = (x y) 2 ; x 3 3x 2 y + 3xy 2 y 3 = (x y) 3 ; x 4 4x 3 y + 6x 2 y 2 4xy 3 + y 4 = (x y) 4 ; x 5 5x 4 y + 10x 3 y 2 10x 2 y 3 + 5xy 4 y 5 = (x y) 5 ; Other factorization formulae: x 3 + y 3 + z 3 3xyz = (x + y + z)(x 2 + y 2 + z 2 xy xz yz). (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2xz + 2yz. (x y 2 1)(x y 2 2) = (x 1 x 2 + y 1 y 2 ) 2 + (x 1 y 2 y 1 x 2 ) Polynomials in one variable x. Completing the square to write a quadratic in vertex form. Consider x 2 + 2xy + y 2 = (x + y) 2. Application:: Writing ax 2 + bx + c = a(x h) 2 + k for some numbers h, k. Example: Example: 3x x + 48 = 3(x x + 16) = 3(x x ) = 3[(x + 5) 2 9]. Why is this useful? a) Solving quadratic equations: 3x x + 48 = 0. Solution:

5 3x x + 48 = 0 becomes 3[(x + 5) 2 9] = 0. Using the difference of two squares this becomes: 3(x + 5 3)(x ) = 0 so 3(x + 2)(x + 8) = 0 so x = 2 or x = 8. b) Find all integers n such that n 2 + 6n + 16 is the square of an integer number. Solution: We are asked to find integer solution for n 2 + 6n + 16 = m 2 But n 2 + 6n + 16 = n n = (n + 3) = m 2 becomes (n + 3) 2 m 2 = 7. Using the difference of squares: (n m)(n + 3 m) = 7 1 = 7 ( 1). We solve simultaneous equations: n m = 7 and n + 3 m = 1. Then 2n + 6 = 6 so n = 6. Or n m = 7 and n + 3 m = 1 so 2n + 6 = 6 so n = Polynomials in one variable x. Division with remainder. Let a(x) and b(x) be two polynomials. Then there exists a unique pair of polynomials q(x) (the quotient) and r(x) (the remainder) for which a(x) = b(x)q(x) + r(x) and degree r(x) < degree q(x). This is similar to the division algorithm for integers. The greatest common divisor d(x) of two polynomials a(x) and b(x) is a polynomial of largest degree which divides exactly into both a(x) and b(x). It can be found by repeated division (Euclid s algorithm). Example: Find the greatest common divisor of x 3 +x 2 3x 6 and x 3 3x 2. By division we get x 3 + x 2 3x 6 = (x 3 3x 2) + (x 2 4) x 3 3x 2 = (x 2 4) x + (x 2) x 2 4 = (x 2) (x + 2) + 0 so x 2 divides x 2 4 and hence also x 3 3x 2 and hence also x 3 +x 2 3x 6. The greatest common divisor is x 2. The Remainder Theorem: Let a be a constant. Dividing a polynomial P (x) by (x a) yields the number P (a) as remainder: Proof: P (x) = (x a)q(x) + P (a).

6 6 Indeed, since (x a) has degree 1, the remainder must have degree 0 and hence be a number, let s call it R. Set x = a to get p(a) = r. P (x) = (x a)q(x) + R. So (x a) is a factor of P (x) if and only if P (a) = 0. Example: Factor P (x) = x 3 5x 2 + 3x + 1. Solution: We search for an integer number a such that P (a) = 0. Then a 3 5a 2 + 3a + 1 = 0 so a 3 5a 2 + 3a = 1. Since the left-hand-side is a multiple of a, then a must be a divisor of 1. We try a = 1. Then P (1) = = 0. The remainder theorem implies that P (x) has x 1 as factor. We can divide P (x) by x 1 as follows: We note that for any number k we have x k (x 1) = x k+1 x k and x k (x 1) = x k+1 + x k. So we can try to write P (x) as a sum of terms of these forms: P (x) = x 3 x 2 4x 2 +4x x+1 = x 2 (x 1) 4x(x 1) (x 1) = (x 1)(x 2 4x 1). By completing the square: P (x) = (x 1)(x 2 4x + 4 5) = (x 1)[(x 2) 2 5] = (x 1)[(x 2) ] = (x 1)(x 2 5)(x 2 5).

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