Lecture : Dividing Rational Expressions

Lecture : Dividing Rational Expressions Recall that rational expressions are essentially fractions, the difference being that they have polynomials as their numerators and denominators rather than the integers we are more familiar with. When we divide with those fractions, we invert the divisor and multiply (or as it is commonly known "flip the second fraction" and multiply, or multiply by the reciprocal of the denominator) Ex) 3 10 9 25 becomes 3 25 x 10 9 Which after cancelling becomes 1 5 x 2 3 And The answer is 5 6 With rational expressions we follow the same steps, we invert the divisor and multiply. Here is an example: (5x² + 10x) (x³- 4x) (x² - 1) (5x²+ 10x + 5) First we will invert the second fraction and change the operation to multiplication (5x² + 10x) (5x²+ 10x + 5) x (x³- 4x) (x² - 1) after factoring we will have: 5x(x + 2) 5(x + 1)(x + 1) x x(x - 2)(x + 2) (x - 1)(x + 1) Notice x , (x + 1) and (x + 2) occur in both the numerator and the denominator: 5x(x + 2) 5(x + 1)(x + 1) x x(x - 2)(x + 2) (x - 1)(x + 1) So we can cancel them and get: 5 5(x + 1) x (x - 2) (x - 1) Now we multiply the numerators, and multiply the denominators to get our answer: 25(x + 1) (x - 2)(x - 1) Divide each of the following. 1. (5x² + 7x + 2) (12x² - 19x - 21) 2. (x² + 10x) (x² - 1) (8x² + 13x + 5) (6x² + x - 35) (x³+ 1000) (5x²+ 6x + 1) 3. (20x² + 64x - 21) (6x - 21) (4x²- 49) #1 solution (5x² + 7x + 2) (12x² - 19x - 21) (8x² + 13x + 5) (6x² + x - 35) First we will invert the second fraction and change the operation to multiplication (5x² + 7x + 2) (6x² + x - 35) x (12x² - 19x - 21) (8x² + 13x + 5) after factoring we will have: (5x + 2)(x + 1) (3x - 7)(2x + 5) x (4x + 3)(3x - 7) (x + 1)(8x + 5) Notice that (x + 1) and (3x - 7) occur in both the numerator and the denominator: (5x + 2)(x + 1) (3x - 7)(2x + 5) x (4x + 3)(3x - 7) (x + 1)(8x + 5) So we can cancel them and have: (5x + 2) (2x + 5) x (4x + 3) (8x + 5) There are no more common factors so we multiply the numerators, and multiply the denominators to get our answer: (5x + 2)(2x + 5) (4x + 3)(8x + 5) #2 solution (x² + 10x) (x² - 1) (x³+ 1000) (5x²+ 6x + 1) First we will invert the second fraction and change the operation to multiplication (x² + 10x) (5x²+ 6x + 1) x (x² - 1) (x³+ 1000) after factoring we will have: x (x + 10 ) (5x + 1)(x + 1) x (x + 1)(x - 1) (x + 10)(x² - 10x + 100) Since (x + 10) and (x + 1) occurs in both the numerator and the denominator, we can cancel them: x (x + 10 ) (5x + 1)(x + 1) x (x + 1)(x - 1) (x + 10 )(x² - 10x + 100) = x (5x + 1) x (x - 1) (x² - 10x + 100) There are no more common factors so we multiply the numerators, and multiply the denominators to get our answer: x(5x + 1) (x - 1)(x² - 10x + 100) #3 solution 3. (20x² + 64x - 21) (6x - 21) (4x²- 49) Note that (4x² - 49) does not have a denominator, we may put it over 1, so that it is in fraction form. Now we will invert the second fraction and change the operation to multiplication (20x² + 64x 1 x 21) (4x²- 49) (6x - 21) after factoring we will have: (2x + 7)(10x - 3) 1 x 3(2x - 7) (2x + 7)(2x - 7) Since (2x + 7) occurs in both the numerator and the denominator, we can cancel it and have: (10x - 3) 1 x 3(2x - 7) (2x - 7) There are no more common factors so we multiply the numerators, and multiply the denominators to get our answer: (10x - 3) 3(2x - 7)²