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.

4. 4 Simplifying Rational Expressions

5. 5

6. 6 Simplify the following rational expressions. Solution: a. Example 1 – Simplifying rational expressions Factor the numerator and denominator. Divide out any common factors.

7. 7 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

8. 8 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

9. 9 Long Division of Polynomials

10. 10 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.

11. 11 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.

12. 12 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

13. 13 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

14. 14 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

15. 15 Example 5 – Solution Therefore, The remainder remains over the divisor. cont’d