Appendix: Synthetic Division AP Learning Objectives In this section, we will learn how to: 1. Divide polynomials using synthetic division. Synthetic division is a short form of long division with polynomials. We will consider synthetic division only for those cases in which the divisor is of the form x + k, where k is a constant. Let s begin by looking over an example of long division with polynomials. x x + 4 x + ) x + 7x x 4 x + 9x x x x 6x 4x 4 4x + 1 16 We can rewrite the problem without showing the variable since the variable is written in descending powers and similar terms are in alignment. It looks like this: _ 4 1 + ) 7 4 () + 9 ( ) ( ) 6 4 ( 4) (4) 1 16 We have used parentheses to enclose the numbers that are repetitions of the numbers above them. We can compress the problem by eliminating all repetitions except the first one: _ 4 1 + ) 7 4 9 6 1 4 16 The top line is the same as the first three terms of the bottom line, so we eliminate the top line. Also, the 1 that was the coefficient of x in the original problem can be eliminated since we will consider only division problems where the divisor is of the form x + k. The following is the most compact form of the original division problem: _ +) 7 4 9 6 1 4 16 AP-1
AP- APPENDIX Synthetic Division If we check over the problem, we find that the first term in the bottom row is exactly the same as the first term in the top row and it always will be in problems of this type. Also, the last three terms in the bottom row come from multiplication by + and then subtraction. We can get an equivalent result by multiplying by and adding. The problem would then look like this: 7 4 g 9 6 1 4 16 We have used the brackets and to separate the divisor and the remainder. This last expression is synthetic division. It is an easy process to remember. Simply change the sign of the constant term in the divisor, then bring down the first term of the dividend. The process is then just a series of multiplications and additions, as indicated in the following diagram by the arrows: 7 4 g 9 6 1 } Add Multiply 4 16 The last term of the bottom row is always the remainder. The numbers to the left of this term are the coefficients of the terms for the quotient. Here are some additional examples of synthetic division with polynomials. EXAMPLE 1 888n 8888n Divide x 4 x + 4x 6x + by x. SOLUTION To begin, we change the sign of the constant term in the divisor to get + and write this on the left, followed by the coefficients of the dividend. Next, we bring the left-most coefficient straight down. Here is how it looks: + 1 4 6 g 1 Now we multiply the 1 in the bottom line by, and write the result below the next coefficient. Then we add to obtain 0. + 1 4 6 g 1 0 We repeat this procedure with each new number in the bottom row. Here is the result in individual steps: + 1 4 6 g 0 1 0 4 + 1 4 6 g 0 8 1 0 4 888n + 1 4 6 g 0 8 4 1 0 4 6
APPENDIX Synthetic Division AP- From the last line we have the answer. The first term in the quotient will always have a degree that is one less than the degree of the leading term in the dividend. In this case, the quotient will begin with an x term. 1x + 0x 6 + 4x + + x Further simplifying the result, we get: x 6 + 4x + + x x Divide: 4x + 5 EXAMPLE. x + 4 SOLUTION Since we cannot skip any powers of the variable in the dividend x 4x + 5, we rewrite it as x + 0x 4x + 5 and proceed as we did in Example 1: 4 0 4 5 g 1 48 176 1 44 171 From the synthetic division, we have x 4x + 5 = x x + 4 171 1x + 44 + x + 4 x Divide: 1 EXAMPLE x 1. SOLUTION Writing the numerator as x + 0x + 0x 1 and using synthetic division, we have +1 1 0 0 1 g 1 1 1 1 1 1 0 which indicates x 1 x 1 = x + x + 1 Getting Ready for Class After reading through the preceding section, respond in your own words and in complete sentences. A. When can synthetic division be used? B. How do we know what power of the variable the quotient begins with? C. What number goes in the top left bracket? D. How do we handle missing powers of the variable in the dividend?
AP-4 Problem Set Appendix Divide using the long division method. x 1. 5x 7. x + 4x 8 x + x x. x 4x + 5 x 4. 5x + x 1 x + 1 x a 5. 4 a + 5 6. a 4 + a 1 a a + y 7. _ 4 16 y 8. _ y 4 81 y Divide using synthetic division. 9. x 5x + 6 10. x + 8x 1 x + x x 11. 4x + 1 1. 4x x 6 x 1 x + 1 1. x + x + x + 4 14. x x x 4 x x x 15. x + x + 5 x 16. 5x + x + x x x 17. + x 18. x x + 1 x 1 x 5 19. x4 + x + 1 0. x4 x + 1 x + 4 x 4 1. x5 x 4 + x x x + 1 x. 5 x 4 + x x + x + 1 x x +. x + x + 1 4. x + x + 1 x 1 x + 1 5. x4 1 x + 1 6. x4 + 1 x 1 7. x 1 x 1 8. x 1 x + 1
APPENDIX Synthetic Division AP-5 Applying the Concepts 9. The Remainder Theorem Find P( ) if P(x) = x 5x + 6. Compare it with the remainder in Problem 9. 0. The Remainder Theorem The remainder theorem of algebra states that if a polynomial, P(x), is divided by x a, then the remainder is P(a). Verify the remainder theorem by showing that when P(x) = x x + is divided by x the remainder is 5, and that P() = 5. Learning Objectives Assessment The following problems can be used to help assess if you have successfully met the learning objectives for this section. 1. Use synthetic division to divide x + 8 _ x +. a. x + 4 b. x x + 4 c. x 16 + x 4 + x + d. x + x + 4 x. Divide _ x + 6 using synthetic division. x + 8 a. x + 1 + x + b. x + 1 x + c. x + d. x +