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Product Rule for Radicals If n a and n b are real numbers and n is a natural number, then Simplifying Radicals Objective: To simplify radical expressions using the product and quotient rules. n a ( b) = n n ab . In other words, the product of two radicals is the radical of the product. Product Rule Practice ( 3) 23 ( t ) 7x ( 2 y ) 3 y ( 6 yz ) 5a ( 4ab ) SOLUTIONS ( 3 ) = 30 23 ( t ) = 23t 7 x ( 2 y ) = 14 xy 3 y ( 6 yz ) = 18 y z 5a ( 4ab ) = cannot multiply a ) 10 a) 10 b) b) c) 3 d) 4 e) 3 2 6 4 5 Quotient Rule for Radicals If n a and n b are real numbers and c) 3 d) 4 e) 3 6 n In other words, the quotient of two radicals is the radical of the quotient. 4 4 3 5 Quotient Rule Practice f) 16 49 g) 3 216 27 b ≠ 0, and n is a natural number, then n a a = n . b b 2 3 h) 3 − 8 125 i) − 4 625 16 1 SOLUTIONS 16 4 = 49 7 f) g) 3 h) 3 8 2 =− 125 5 i) − 4 625 5 =− 16 2 Rules for Simplifying Radicals 2. 3. 4. z For both the product and quotient properties, you must have the same index! 216 6 = =2 27 3 − 1. Special Note No factor under the radical has a higher power than the root. No fractions under radical. N radical No di l iin denominator. d i t No common factors between root and exponents in radical. Practice Simplifying Radicals Removing Perfect-Square Factors z Re-write the radicand as a product of perfect-square factors times any remaining factors. For example: 75 = 25 ( 3) = 5 3 SOLUTIONS i ) 48 i ) 48 = 16i3 = 4 3 j ) − 24 j ) − 24 = − 4i6 = −2 6 k ) − 5 300 k ) − 5 300 = −5 100 00i3 = −5(10) 5( 0) 3 = −50 3 l ) 18 m) 3 24 l ) 18 = 9i2 = 3 2 m) 3 24 = 3 8i3 = 2 3 3 2 Practice Simplifying Radicals SOLUTIONS n) −150 o) 3 n) −150 = not real −250 o) 3 −250 = 3 −125i2 = −5 3 2 p ) − 4 1250 p ) − 4 1250 = − 4 625i2 = −5 4 2 q ) 5 128 q) 5 128 = 5 32i4 = 2 5 4 Removing Variable Factors z Practice with Variables Divide variable exponents by the root and leave any remainder under the radical. For example: 11 2 1 2 x = x =x x =x 11 5 r ) 18m 2 s) 256 z12 5 x t ) − 200 p13 SOLUTIONS r ) 18m 2 = 9i2im 2 = 3m 2 s) 256 z12 = 16 z 6 t ) − 200 p13 = − 100i2i p12 i p = −10 p 6 2 p Practice with Variables u ) − 3 216 y15 x 6 z 3 v) 23k 9 p14 w) − 4 32k 5 m10 3 SOLUTIONS u ) − 3 216 y15 x 6 z 3 = −6 y 5 x 2 z v) 23k 9 p14 = 23k 8 ik i p14 = k 4 p 7 23k w) − 4 32k 5 m10 = − 4 16i2ik 4 ik im8 im 2 = −2km 2 4 2m 2 k 4