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Example 1: Factor 8x3 + 27 A is 2x and B is 3 A3 + B3 = (A + B)(A2 - AB + B2) = (2x + 3 )((2x)2 - 2x·3 + 32) = (2x + 3 )(4x2 - 6x + 9) Once again, study each line so you can see what we did. We'll dispense with the colour coding from here on. Example 2: (a simpler one) Factor x3 - 125 Here A = x and B = 5 A3 - B3 = (A - B)(A2 + AB + B2) x3 - 125 = (x - 5)(x2 + 5x + 25) Example 3: 8p9q6 + 125p12q3 You should always write out what A and B are: A = 2p3q2 B = 5p4q1 and we'll drop the 1 A3 + B3 = (A + B) · (A2 - AB + B2) 8p9q6 + 125p12q3 = (2p3q2 + 5p4q) · ((2p3q2)2 - 2p3q2·5p4q + (5p4q)2) = (2p3q2 + 5p4q)(4p6q4 - 10p7q3 + 25p8q2) Both factors have common factors, which must be removed: = p3q(2q + 5p)·p6q2(4q2 - 10pq + 25p2) = p9q3(2q + 5p)(4q2 - 10pq + 25p2) As you learned in Math 10C, it's always best to do the common factor first so you don't forget. Here is the example done over again, where we remove the common factor first. 8p9q6 + 125p12q3 = p9q3(8q3 + 125p3) The second bracket contains the sum of cubes. A = 2q and B = 5p = p9q3(2q + 5p)((2q)2 - 10pq + (5p)2) = p9q3(2q + 5p)(4q2 - 10pq + 25p2) ... which of course is the same answer as above. This way is easier too! Example 4: x5 + 1000x2 This one is clearly not a sum of cubes. However, there is a common factor: x5 + 1000x2 = x2(x3 + 1000) Now we can factor the sum of cubes, leaving the common factor in front: A = x B = 10 = x2(x + 10)(x2 - 10x + 100) |