Sect 3.2 Synthetic Division

94 Objective 1: Sect 3.2 Synthetic Division Division Algorithm Recall that when dividing two numbers, we can check our answer by the whole number (quotient) times the divisor plus the remainder. This should be equal to the original number we were dividing into (dividend). Division Problem: Check: 217 R 13 7825 = 36 217 + 13 36) 7825 Dividend = Divisor Quotient + Remainder Ans: 217 13 36 Quotient Remainder Divisor We can extend this idea to polynomials: Division Problem: Check: 5x 4 R 11 5x 2 + 6x + 3 = (x + 2) (5x 4) + 11 x + 2) 5x 2 + 6x+ 3 Dividend = Divisor Quotient + Remainder Ans: 5x 4 + 11 x+2 Quotient + Remainder Divisor P(x) = D(x) Q(x) + R(x) We call the "check" the Division Algorithm since it relates the relationship between division and multiplication. Division Algorithm Let P(x) (Dividend) and D(x) (Divisor) be two polynomials such that degree of D(x) > 0 and the degree of D(x) is less than P(x). Then, there exists unique polynomials Q(x) (Quotient) and R(x) (Remainder) such that: P(x) = D(x) Q(x) + R(x) where R(x) = 0 or the degree of R(x) is less than the degree of D(x). Objective 2: Understanding and Using Synthetic Division. When dividing a polynomial by binomial in the form of x c, we can use a process called synthetic division. The advantages of synthetic division is that it is easier to perform than long division. The disadvantages are that can only be used when the divisor is in the form x c (binomial of degree

one, coefficient of the variable term is one). To see where synthetic division comes from, let s look at a long division. Perform the indicated operation: Ex. 1 (x 2 + 11x 19) (x 4) x 2 x = x, subtract x(x 4) which is x + 15 the same as adding x 2 + 4x. x 4 x 2 + 11x 19 Bring down the 19. 15x x = 15, x 2 + 4x subtract 15(x 4) or add 15x + 60. 15x 19 Put the remainder over the divisor. 15x + 60 So, the answer is x + 15 + 41 x 4. 41 The important information in the problem are the coefficients. So, let s drop the variables and only list the variables. x + 15 1 + 15 x 4 x 2 + 11x 19 1 4 1 + 11 19 x 2 + 4x 1 + 4 ( 4 1) 15x 19 15 19 15x + 60 15 + 60 ( 4 15) 41 41 The + 4 came from subtracting a negative 4 times 1. The + 60 came from subtracting a negative 4 times 15. But subtracting a negative is the same as adding a positive, so to make this easier, we will change the sign of the 4 in the divisor. Also, we do not need to keep track of the 1 and 15 so we will drop them. Lastly, we will drop the plus signs. Drop the numbers in bold, change the sign of 4 to 4 1 + 15 1 15 1 4 1 + 11 19 4 1 11 19 1 + 4 4 (4 1) 15 19 15 19 15 + 60 60 (4 15) 41 41 95

So, we take 4 times 1, add to 11 to get 15. Then, we take 4 times 15 and add 19. The answer is the remainder and the 1 and 15 are the coefficients. To make this easier to read, we will reorganize the algorithm. Delete the coefficients of the divisor, and make the 1, 15, and 41 along the diagonal the bottom row. 1 15 4 1 11 19 4 1 11 19 4 4 60 15 19 1 15 41 60 41 Now, let s run through the algorithm: (x 2 + 11x 19) (x 4) Since the divisor is in the form of x c, c is 4. We write 4 in the upper left. Next, list the coefficients of the dividend in the upper right. Bring down the first coefficient (1), multiply it by 4 and add the result to the second coefficient (11) to get 15. Finally, multiply 15 by 4 and add the result to the third coefficient ( 19) to get 41. The last number, 41, is the divisor, the other numbers are the coefficients of the quotient. 4 1 11 19 4 60 1 4 1 15 4 15 41 Note the degree of the quotient will always be one less than the degree of the dividend whenever we are using synthetic division. Since the degree of the dividend was 2, the degree of the quotient will be one. It is also important that the polynomials be listed in order of descending powers before starting the synthetic division. Finally, if there are terms missing, we will need to use zeros as placeholders in the synthetic division. Use synthetic division to divide: Ex. 2 (x 3 7x 2 13x + 3) (x + 2) Since x + 2 = x ( 2), c = 2. 2 1 7 13 3 Bring down the 1. 2 1 = 2. 2 18 10 List it under 7 and add to get 1 9 5 7 9. 2( 9) = 18. List it under 13 and add to get 5. 2 5 = 10. List it under 3 and add to get 7. 1 is the coefficient of the x 2, 9 the coefficient of x, five is the constant term and 7 is the remainder. So, we get x 2 9x + 5 96 7 x+2.

97 Ex. 3 (6x 3 3x 2 45x 9) (x 3) c = 3 3 6 3 45 9 Bring down the 6. 3 6 = 18. 18 45 0 List it under 3 and add to get 6 15 0 9 15. 3(15) = 45. List it under 45 and add to get 0. 3 0 = 0. List it under 9 and add to get 9. 6 is the coefficient of the x 2, 15 the coefficient of x, zero is the constant term and 9 is the remainder. So our answer is 6x 2 + 15x 9 x 3. Ex. 4 (x 3 64) (x 4) Since the dividend has no x 2 and x terms, we will need to use zeros as place holders in the synthetic division. c = 4 4 1 0 0 64 Bring down the 1. 4 1 = 4. 4 16 64 List it under the first 0 and add to get 1 4 16 0 4. 4(4) = 16. List it under the second 0 and add to get 16. 4 16 = 64. List it under 64 and add to get 0. 1 is the coefficient of the x 2, 4 the coefficient of x, sixteen is the constant term and 0 is the remainder. So our answer is x 2 + 4x + 16. Ex. 5 (8r 4 2 + 12r + 10r 3 5r 2 ) (r + 1 2 ) First, write the polynomials in order of descending powers: (8r 4 + 10r 3 5r 2 + 12r 2) (r + 1 2 ) Since r + 1 2 = r ( 1 2 ), c = 1 2. 1 2 8 10 5 12 2 4 3 4 8 8 6 8 16 10 So, (8r 4 + 10r 3 5r 2 + 12r 2) (r + 1 2 ) = 8r3 + 6r 2 8r + 16 10 r+ 1 2.

If we were to multiply both sides of the result from the above example by (r + 1 ), we would get: 2 8r 4 + 10r 3 5r 2 + 12r 2 = (r + 1 2 )(8r3 + 6r 2 8r + 16) 10. P(x) = (x c) Q(x) + r This illustrates the special case of the Division Algorithm: Special Case of the Division Algorithm Let P(x) be a polynomial and c be any complex number. Then, there exists unique polynomial Q(x) and a number r such that: P(x) = (x c) Q(x) + r It is important to note that one needs to be dividing by a linear factor for the Special Case of the Division Algorithm to work and to be able to use synthetic division. Ex. 6 (3x 3 2x 2 + 6x + 5) (x 2 + 1) Since the degree of the divisor is not 1, we cannot use synthetic division on this problem. We have to use long division. 3x 3 x 2 = 3x, subtract 3x(x 2 + 1) or add 3x 3 3x, being careful to align like terms. Bring down the 5. 2x2 = 2, subtract 2(x 2 + 1) x 2 or add 2x 2 + 2, being careful to line up like terms. Since the degree of the divisor is higher than what is left, we stop and put the remainder over the divisor. 3x 2 + 3x+7 x 2 +1 x 2 + 1 3x 3 2x 2 + 6x + 5 3x 3 3x 2x 2 + 3x + 5 2x 2 2 3x + 7 98 Objective 3: Using the Remainder Theorem to Evaluate Polynomials.

99 a) Divide f(x) by the x + 3 and then b) find f( 3). Ex. 7 f(x) = 6x 4 + 15x 3 + 28x + 5 a) Here, we can use synthetic division. Since x + 3 = x ( 3), c = 3. Also, the dividend has no x 2, we will need to a zero as a place holder: 3 6 15 0 28 5 18 9 27 3 6 3 9 1 2 So, the answer is 6x 3 3x 2 + 9x + 1 + 2 x+3. b) f( 3) = 6( 3) 4 + 15( 3) 3 + 28( 3) + 5 = 6(81) + 15( 27) + 28( 3) + 5 = 486 405 84 + 5 = 2 Notice that the answer is the same as the remainder. To see why f( 3) is equal to remainder, recall that the dividend is equal to the quotient times the divisor plus the remainder. So, f(x) = (6x 3 3x 2 + 9x + 1)(x + 3) + 2 = 6x 4 + 15x 3 + 28x + 5 Thus, f( 3) = (stuff)( 3 + 3) + 2 = (stuff)(0) + 2 = 0 + 2 = 2. This is known as the remainder theorem. The Remainder Theorem The remainder obtained by dividing the polynomial P(x) by x c is P(c). Find P(2) by synthetic division: Ex. 8 P(x) = x 4 + 7x 3 + 11x 2 7x 12 Here, c = 2, so 2 1 7 11 7 12 2 18 58 102 1 9 29 51 90 Since the remainder is 90, then P(2) = 90.

100 Objective 4: Using the Remainder Theorem to Test Potential Zeros. A zero of a polynomial P(x) is a number c such that P(c) = 0. If c is a real number, then (c, 0) is an x-intercept of P(x). We can use the remainder theorem to quick determine if c is a zero of P(x). If the remainder of P(x) (x c) is zero, then c is a zero and we also have the quotient of P(x) divided by (x c). Determine whether the given number k is a zero of f(x). Ex. 9 f(x) = 8x 3 + 14x 2 19x 30; k = 2 Using synthetic division, we get: 2 8 14 19 30 16 4 30 8 2 15 0 So, since the remainder is zero, 2 is a zero of f(x). In fact, we can rewrite f(x) as: 8x 3 + 14x 2 19x 30 = (x + 2)(8x 2 2x 15) Ex. 10 f(x) = x 4 6x 3 + 2x 2 + 57x 90; k = 3 Using synthetic division, we get: 3 1 6 2 57 90 3 9 21 108 1 3 7 36 18 So, 3 is not a zero of f(x). Ex. 11 f(x) = 4x 3 21x 2 + 72x 65; k = 2 3i Using synthetic division, we get: 2 3i 4 21 72 65 8 12i 62 + 15i 65 4 13 12i 10 + 15i 0 So, since the remainder is zero, 2 3i is a zero of f(x). We shall later that if a complex number is a zero, its complex conjugate will also be a zero.