Basic Algebraic Identities with Examples

Some basic algebraic identities

1. Algebraic Identities The Square of a Sum ( x + y ) 2 = ( x + y ) ( x + y ) = x ( x + y ) + y ( x + y ) = x 2 + xy + xy + y 2 = x 2 + 2 xy + y 2 ( x + y ) 2 Problem 1 Expand the following: a. (2 a + 3 b ) 2 b. ( m + 5 n ) 2 Solution a. (2 a + 3 b ) 2 = (2 a ) 2 + 2(2 a × 3 b ) + (3 b ) 2 = 4 a 2 + 12 ab + 9 b 2 b. ( m + 5 n ) 2 = m 2 + 10 mn + 25 n 2

2. Algebraic Identities The Square of a Difference ( x – y ) 2 = ( x – y ) ( x – y ) = x ( x – y ) – y ( x – y ) = x 2 – xy – xy + y 2 = x 2 – 2 xy + y 2 ( x – y ) 2 Problem 2 Expand the following: a. (3 p – q ) 2 b. ( y – 2 z ) 2 Solution a. (3 p – q ) 2 = (3 p ) 2 – 2(3 p × q ) + ( q ) 2 = 9 p 2 – 6 pq + q 2 b. ( y – 2 z ) 2 = y 2 – 4 yz + 4 z 2

3. Algebraic Identities Difference of Two Squares ( x + y ) ( x – y ) = x ( x – y ) + y ( x – y ) = x 2 – xy + xy – y 2 = x 2 – y 2 x 2 – y 2 x 2 – y 2 = ( x + y ) ( x – y ) Problem 3 Expand the following: a. (3 a + 2 b )(3 a – 2 b ) b. (5 x + 4 y )(5 x – 4 y ) Solution a. (3 a + 2 b )(3 a – 2 b ) = (3 a ) 2 – (2 b ) 2 = 9 a 2 – 4 b 2 b. (5 x + 4 y )(5 x – 4 y ) = 25 x 2 – 16 y 2

4. Algebraic Identities Difference of Two Squares Problem 4 Evaluate the following using difference of two squares: (a) 8.9 2 – 1.1 2 (b) 98 2 – 2 2 x 2 – y 2 = ( x + y ) ( x – y ) Solution (a) 8.9 2 – 1.1 2 = (8.9 + 1.1) (8.9 – 1.1) = 10.0 × = 78 7.8 (b) 98 2 – 2 2 = (98 + 2) (98 – 2) = 100 96 × = 9,600

5. Algebraic Identities Sum of Two Cubes x 3 + y 3 ( x + y ) ( x 2 – xy + y 2 ) = x ( x 2 – xy + y 2 ) + y ( x 2 – xy + y 2 ) = x 3 – x 2 y + xy 2 + x 2 y – xy 2 + y 3 – x 2 y + x 2 y + xy 2 – xy 2 = x 3 + y 3 ⸫ x 3 + y 3 = ( x + y ) ( x 2 – xy + y 2 ) Problem 5 Simplify . x 3 + y 3 x + y x 3 + y 3 x + y = ( x + y ) ( x 2 – xy + y 2 ) x + y = x 2 – xy + y 2

6. Algebraic Identities Difference of Two Cubes x 3 – y 3 ( x – y ) ( x 2 + xy + y 2 ) = x ( x 2 + xy + y 2 ) – y ( x 2 + xy + y 2 ) = x 3 + x 2 y + xy 2 – x 2 y – xy 2 – y 3 + x 2 y – x 2 y + xy 2 – xy 2 = x 3 – y 3 ⸫ x 3 – y 3 = ( x – y ) ( x 2 + xy + y 2 ) Problem 6 Simplify . x 3 – y 3 x 2 + xy + y 2 x 3 – y 3 = x 2 + xy + y 2 ( x – y ) ( x 2 + xy + y 2 ) x 2 + xy + y 2 = x – y