1. Copyright © Cengage Learning. All rights reserved. 7 Rational Functions

2. Copyright © Cengage Learning. All rights reserved. 7.2 Simplifying Rational Expressions

3. 3 Objectives  Simplify rational expressions using factoring.  Divide polynomials using long division.  Divide polynomials using synthetic division.

4. 4 Simplifying Rational Expressions

5. 5 Simplifying rational expressions is the same as simplifying numeric fractions. The numerator and denominator must both include multiplication for a factor to divide out and thus reduce the fraction. To work with rational expressions, we will need to focus on each step of the process to properly reduce the expression.

6. 6 Simplifying Rational Expressions

7. 7 Simplify the following rational expressions. Solution: a. Example 1 – Simplifying rational expressions Factor the numerator and denominator. Divide out any common factors.

8. 8 Example 1(b) – Solution Already factored, so divide out common factors. When all the factors in the numerator or denominator divide out, a 1 is left. Reduce. cont’d

9. 9 Example 1(c) – Solution Divide out common factors. Note that the remaining x does not reduce because it is not multiplied, but subtracted. cont’d

10. 10 Long Division of Polynomials

11. 11 Long Division of Polynomials When we divide a polynomial by another polynomial, we can use long division. Long division of polynomials is basically the same process as long division with numbers. We will start by using long division to divide 458 by 6. 6 does not divide into 4, so we divide into 45 first. 6 divides into 45 seven times, so 7 goes above the 5, and we multiply 6(7) = 42. Subtract and bring down the next digit (8). 6 divides into 38 six times, so the 6 goes above the 8, and we multiply 6(6) = 36 Subtract. The remainder is 2.

12. 12 Long Division of Polynomials From this long division, we get We can check this answer by multiplying it by 6. This same division process can be used with polynomials. When the divisor has a higher degree than the remainder, we can stop the process. The remainder will be written over the original divisor.

13. 13 Example 5 – Dividing a polynomial by a polynomial a. Divide 3x 2 + 17x + 20 by x + 4. b. Divide 10x 2 + 7x – 19 by –x + 3. Solution: a. Divide the first term 3x 2 by x. Multiply x + 4 by 3x and subtract. To subtract, distribute the negative sign and combine like terms. Bring down the next term, 20

14. 14 Example 5 – Solution Continue dividing 5x by x. Multiply x + 4 by 5 and subtract. Therefore, Distribute the negative sign and combine like terms The remainder is zero, so we are done cont’d

15. 15 Example 5 – Solution Divide the first term 10x 2 by 2x. Multiply 2x + 3 by 5x and subtract. Bring down the next term. b. Divide the –8x by 2x. Multiply 2x + 3 by –4 and subtract. The divisor, 2x + 3 has a higher degree than the remainder, –7, so we can stop here. cont’d

16. 16 Example 5 – Solution Therefore, The remainder remains over the divisor. cont’d

17. 17 Synthetic Division

18. 18 Synthetic Division Using long division requires that we keep track of a lot of details along the way. Writing out some of the steps used in long division some times is not necessary. Simplifying the procedure to keep track of only the most important steps results in a shorter method than long division. This method is called synthetic division. Synthetic division is a simplified method of long division that uses only the coefficients of each term and not the variables.

19. 19 Synthetic Division This method works only for single-variable polynomials that are being divided by a binomial of the form x – c. Let’s compare the long division in Example 5a with the synthetic division of the same polynomials. Divide 3x 2 + 17x + 20 by x + 4

20. 20 Synthetic Division The basic setup is similar but without the variables. Synthetic division works for divisors of the from x – c, and x + 4 can be written in this form as x –(–4). Therefore, we start with a –4 representing the divisor.

21. 21 Synthetic Division In synthetic division, the leading coefficient comes down to the bottom row in the first column. Multiply this by the divisor, –4. The result goes into the second row and next column. By changing the sign of 4 in the divisor, we accounted for the subtraction in the long division. We add the values in the second column to get the next value in the bottom row.

22. 22 Synthetic Division Multiply the 5 in the bottom row by the divisor, –4, and the result goes in the second row and next column to the right. Add the values in the next column, and we get zero in the bottom row. When all the columns have been filled, we are done. The zero in the bottom row and last column means that there is no remainder. Any nonzero number in this position would be the remainder.

23. 23 Synthetic Division Once we are done with the division we can write the result using the numbers in the last row. Each number in the last row is a coefficient for a term in the final answer.

24. 24 Synthetic Division The first number will be the coefficient of the first term that will have degree one less than the degree of the dividend. The dividend had a degree of 2, so we start with a first degree term.

25. 25 Example 7 – Using synthetic division a. Use synthetic division to divide 3x 3 – 10x 2 – 13x + 20 by x – 4. b. Use synthetic division to divide 5x 4 + 2x 3 – x – 50 by x + 2.

26. 26 Example 7(a) – Solution Since the divisor, x – 4 is of the form x – c, we know that c is 4. The dividend’s coefficients are 3, –10, –13, and 20. Write out the coefficients in order. Draw a line and bring down the first coefficient. Multiply 4 by 3 and write the product under the next coefficient.

27. 27 Example 7(a) – Solution Add the –10 and 12 and write the sum below. Multiply 4 by 2 and write the product under the next coefficient. Add the –13 and 8 and write the sum below. Multiply 4 by –5 and write the product under the next coefficient. Add the –20 and 20 and write the sum below. cont’d

28. 28 We have completed each column, so we have finished the synthetic division. Each number below the line is a coefficient for a term in the final answer. The first number will be the coefficient of the first term that will have degree one less than the degree of the dividend. The dividend had a degree of 3, so start with a term with degree 2. These coefficients result in the quotient. 3x 2 + 2x – 5 The zero on the end is the remainder, so this answer does not have a remainder. Example 7(a) – Solution cont’d

29. 29 The divisor x + 2 can be written in the form x – c as x –(–2), so c = –2. The dividend is missing a second-degree term, so we will write zero as its coefficient. Start by writing c and the coefficients in order. Example 7(b) – Solution cont’d

30. 30 The last number in the bottom row is 16, so that is our remainder. Therefore, we have the following quotient. Example 7(b) – Solution cont’d