1 Topic Dividing Rational Expressions Dividing Rational Expressions

2 Lesson California Standard: 13.0 Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll divide rational expressions by factoring and cancelling. Dividing Rational Expressions Topic Key words: rational reciprocal common factor

3 Lesson Dividing by rational expressions is a lot like multiplying — you just have to do an extra step first. Topic That extra step is finding the reciprocal. Dividing Rational Expressions

4 Lesson Dividing is the Same as Multiplying by the Reciprocal Topic Given any nonzero expressions m, c, b, and v : m c ÷ = = b v v b m c mv cb That is, to divide by, multiply by the reciprocal of. m c b v b v Dividing Rational Expressions

5 Lesson Dividing is the Same as Multiplying by the Reciprocal Topic You can extend this concept to the division of any rational expression. The question you’re trying to answer is… “How many times does go into 10?” …or “How many halves are in 10?” 1 2 Suppose you pick a number such as 10 and divide by. 1 2 Dividing Rational Expressions

6 10 ÷ = × 2 = ÷ = × 3 = ÷ = × 4 = ÷ = 10 n 10 × n = 10 n Lesson Dividing is the Same as Multiplying by the Reciprocal Topic Division Equivalent to So, 10 divided by a fraction is equivalent to 10 multiplied by the reciprocal of that fraction. Dividing Rational Expressions

7 Lesson Dividing is the Same as Multiplying by the Reciprocal Topic Dividing anything by a rational expression is the same as multiplying by the reciprocal of that expression. So you can always rewrite an expression a ÷ b in the form a = (where b is any nonzero expression). 1 b a b As always, you should cancel any common factors in your answer to give a simplified fraction. Dividing Rational Expressions

8 Example 1 Solution follows… Topic Solution Rewrite the division as multiplication by the reciprocal of the divisor. Simplify ÷ ( k + 5). k 2 – 25 2k2k ÷ ( k + 5) can be written as: k 2 – 25 2k2k ÷ 2k2k k = k 2 – 25 2k2k 1 k + 5 Factor as much as you can: = ( k – 5)( k + 5) 2k2k 1 k + 5 Solution continues… Dividing Rational Expressions

9 Example 1 Topic Solution (continued) Check your answer. Multiply your answer by ( k + 5): Simplify ÷ ( k + 5). k 2 – 25 2k2k Cancel any common factors between the numerators and denominators. = ( k – 5)( k + 5) 2k2k 1 k = k – 5 2k2k = 2k2k ( k – 5)( k + 5) 2k2k k 2 – 25 2k2k k = Dividing Rational Expressions

10 Cancel any common factors between the numerators and denominators. Example 2 Topic Solution Rewrite the division as multiplication by the reciprocal of the divisor. Solution follows… Factor all numerators and denominators. Simplify ÷. m 2 – 4 m 2 – 3 m + 2 2m2m m – 1 m 2 – 4 m 2 – 3 m + 2 m – 1 2m2m = ( m + 2)( m – 2) ( m – 2)( m – 1) m – 1 2m2m = m + 2 2m2m = Dividing Rational Expressions

11 Divide and simplify each expression Lesson Guided Practice Topic Solution follows… 1 b2c2d2b2c2d2 bdc 3 bcd 2 abc ÷ a 2 – 9 a 2 + a – 6 a + 3 a – 2 ÷ a ( b – 2) b + 1 2a2a ( b + 1)( b – 1) ÷ a a + 2 a 2 – a – 6 a 2 – 1 a 2 – 4 a + 3 ÷ x 2 – 5 x – 6 x x – 10 x 2 – 4 x – 5 x 2 – 25 ÷ abc d3d3 b 2 – 3 b a – 3 a + 3 x – 6 x – 2 Dividing Rational Expressions

12 Lesson You Can Divide Long Strings of Expressions At Once Topic Just like multiplication, you can divide any number of rational expressions at once, but it makes a big difference which order you do things in. If there are no parentheses, you always work through the calculation from left to right, so that: a b ÷ c d e f a b = ÷ d c e f a b = d c f e Dividing Rational Expressions

13 Rewrite each division as a multiplication by the reciprocal of the divisor. Example 3 Topic Solution Solution follows… Simplify x x + 6 x x x 2 + x – 2 2 x x ÷ x x + 1 x – 1 ÷.= x x + 6 x x 2 x x x 2 + x – 2 x x + 1 x – 1 ÷= x x + 6 x x 2 x x x 2 + x – 2 x – 1 x x + 1 Solution continues… Dividing Rational Expressions

14 Example 3 Topic Solution (continued) Factor all numerators and denominators. Cancel any common factors between the numerators and denominators. Simplify x x + 6 x x x 2 + x – 2 2 x x ÷ x x + 1 x – 1 ÷.= x x + 6 x x 2 x x x 2 + x – 2 x – 1 x x + 1 = ( x + 2)( x + 3) x ( x + 3) 2 x ( x + 1) ( x + 2)( x – 1) x – 1 ( x + 1)( x + 1) = 2 x Dividing Rational Expressions

15 Lesson You Can Divide Long Strings of Expressions At Once Topic Parentheses override this order of operations, so you need to simplify any expressions in parentheses first: a b = ÷ c f d e a b = c f a b ÷ c d e f Dividing Rational Expressions

16 Divide and simplify each expression Lesson Guided Practice Topic Solution follows… k 2 – 1 2 k 2 – 14 k k k – 6 k 2 – 9 k + 14 ÷ – k k – 2 2 k 2 – 10 k ÷ x 2 – 4 x – 12 2 x 2 – 3 x – 2 – x x x x 2 – 18 x ÷ 6 x 3 – 36 x 2 –2 x x + 4 ÷ ( x – 2) ( x + 3) ( x – 2)( x + 4) ( x + 3) ÷÷ ( x + 4) ( x + 1) x + 3 2x2x 1 x + 1 k 2 – 4 k – 5 k k – 6 – Dividing Rational Expressions

17 Lesson You Can Multiply and Divide at the Same Time Topic Say you have an expression like this to simplify: Again, you work from left to right, and anywhere you get a division, multiply by the reciprocal, so: a b ÷ × c d e f a b c d e f a b × d c e f = Dividing Rational Expressions

18 Rewrite any divisions as multiplications by reciprocals. Example 4 Topic Solution Solution follows… Simplify p 2 + pq – 2 q 2 p 2 – 2 pq – 3 q 2 p 2 + q 2 pq + 2 q 2 × p 2 – 2 pq + q 2 p 2 – 3 pq ÷.= p 2 + pq – 2 q 2 p 2 – 2 pq – 3 q 2 pq + 2 q 2 p 2 + q 2 × p 2 – 2 pq + q 2 p 2 – 3 pq × Factor all numerators and denominators. = p q = ( p + 2 q )( p – q ) ( p – 3 q )( p + q ) ×× ( p – q )( p + q ) q ( p + 2 q ) p ( p – 3 q ) ( p – q )( p – q ) Cancel any common factors Dividing Rational Expressions

19 Example 5 Topic Solution Solution follows… Justify your work. Show that 2 a 2 – 7 a + 3 a a – 21 ÷ 2 a – 2 a + 1 = 2 a 2 t + at – t 2 a 2 t + 12 at – 14 t The question asks you to justify your work, so make sure you can justify all your steps. Start with left-hand side Definition of division 2 a 2 – 7 a + 3 a a – 21 ÷ 2 a 2 t + at – t 2 a 2 t + 12 at – 14 t = 2 a 2 – 7 a + 3 a a – 21 × 2 a 2 t + 12 at – 14 t 2 a 2 t + at – t Solution continues… Dividing Rational Expressions

20 = 2 a 2 – 7 a + 3 a a – 21 × 2 a 2 t + 12 at – 14 t 2 a 2 t + at – t Example 5 Topic Solution (continued) Justify your work. Show that 2 a 2 – 7 a + 3 a a – 21 ÷ 2 a – 2 a + 1 = 2 a 2 t + at – t 2 a 2 t + 12 at – 14 t = (2 a – 1)( a – 3) ( a + 7)( a – 3) × 2 t ( a a – 7) t (2 a 2 + a – 1) = (2 a – 1)( a – 3) ( a + 7)( a – 3) × 2 t ( a + 7)( a – 1) t (2 a – 1)( a + 1) Distributive property Equation carried forward Solution continues… Dividing Rational Expressions

21 = (2 a – 1)( a – 3) ( a + 7)( a – 3) × 2 t ( a + 7)( a – 1) t (2 a – 1)( a + 1) = (2 a – 1) ( a + 1) × t (2 a – 1)( a + 7)( a – 3) t (2 a – 1)( a + 1)( a – 3) Example 5 Topic Solution (continued) Justify your work. Show that 2 a 2 – 7 a + 3 a a – 21 ÷ 2 a – 2 a + 1 = 2 a 2 t + at – t 2 a 2 t + 12 at – 14 t = (2 a – 1) ( a + 1) = 2 a – 2 a + 1 Inverse and identity properties, and distributive property Commutative and associative properties of multiplication Equation carried forward Dividing Rational Expressions

22 Simplify these rational expressions Lesson Guided Practice Topic Solution follows… t 2 – 1 t t – 3 t + 1 t t + 3 ÷ t – 1 1 × a 2 + a – 12 a 2 + a – 2 a a + 4 a a + 1 ÷ a a – 3 a 2 – 2 a – 3 × x x – 14 x 2 – 4 x – 21 x x – 7 x 2 – 6 x – 7 ÷ x x – 3 x 2 – 5 x + 6 × a 2 – 1 a 2 – 4 a 2 – 2 a – 3 a 2 – 3 a – 10 ÷ a 2 – 5 a + 6 – a a + 15 × a + 3 a + 2 x + 1 x – 3 t 2 – 1 a – 1 a + 3 – Dividing Rational Expressions

23 Independent Practice Solution follows… Topic Divide and simplify each expression b 2 + b – 2 k 2 – m 2 2 k 2 + km – m 2 2 k m 2 k km – 2 m 2 ÷ t t – 3 t t t – 3 t 2 – t – 2 ÷ – x 3 – 3 x 2 – 2 x x 2 – 2 x – 3 x 2 – x – 6 x 3 – 2 x 2 – 3 x ÷ b 3 – 4 b b 3 + b b 2 – b – 2 b 4 – 1 ÷ x 2 – 6 x + 8 x 2 – 4 – x 3 + x –2 x x + 16 ÷ y 2 – y – 2 y y – 4 – y + 2 y 2 – 3 y + 2 ÷ k 2 + km – 2 m 2 2 k + 2 m 2 x 2 – 16 x + 32 x 3 – x t – 2 3 x 3 + x 2 x – 3 – y 2 – y – 2 y + 4 – Dividing Rational Expressions

24 Independent Practice Solution follows… Topic Divide and simplify each expression a 3 – 4 a – a a a 2 + a – 2 a 2 – a – 2 ÷ b 2 – 1 b 2 – 2 b – 3 b 2 – 2 b + 1 b 2 – 4 b + 3 ÷ ( m – v ) 2 m 2 – v 2 m 2 – 3 mv + 2 v 2 ( m – 2 v ) 2 ÷ x 2 – 3 x + 2 x 2 + x – 2 – x + 2 x 2 – 3 x – 10 ÷ a 2 – a – 2 a – 1 – 1 m – 2 v m + v – x + 5 Dividing Rational Expressions

25 Independent Practice Solution follows… Topic Divide and simplify each expression x 2 – 5 x – 12 4 x x + 3 x 2 – 16 2 x x + 3 ÷ x 2 – 9 x x – 8 ÷ y + 5 y 2 – 4 y – 5 y y – 5 y + 1 ÷ 1 y 2 – 6 y + 5 ÷ t 2 – t – 6 t t + 9 ÷ ( t 2 – 4) ÷ t + 2 t x – 2 x – 3 1 ( t + 2) 2 Dividing Rational Expressions

26 Independent Practice Solution follows… Topic Simplify these rational expressions k 2 – 5 k + 6 k k – 8 –2 k 2 – 6 k – 4 k 2 – 2 k – 3 k k + 2 k k + 4 ÷ –2 v vw 3 v 2 – 4 vw + w 2 × v 2 – w 2 –2 vw + 3 w 2 ÷ v 3 – vw 2 6 v vw 2 – 2 w 3 –2 v vw – 3 w 2 –4 v vw + 8 w 2 × m mn + n 2 m 2 n – 3 mn 2 ÷÷÷ 4 m mn + n 2 2 m 2 – 5 mn – 3 n 2 2 m mn + n 2 – m mn – 2 n 2 –2 m 2 n + 5 mn 2 m mn + n 2 m 2 n – 3 mn 2 ÷÷÷ 4 m mn + n 2 2 m 2 – 5 mn – 3 n 2 2 m mn + n 2 – m mn – 2 n 2 –2 m 2 n + 5 mn 2 2 m – 5 n 4 m + n – –2 1 ( m + n ) 2 (2 m + n ) 2 (–2 m + 5 n ) (4 m + n )( m – 2 n ) 2 (– m + n ) 2 Dividing Rational Expressions

27 Topic Round Up It’s really important that you can justify your work step by step, because division of rational expressions can involve lots of calculations that look quite similar. Dividing Rational Expressions