page two We want to be able to do a problem like this: (2a3 - 5b2)4 where we aren't just dealing with x and y, but other variables with their own exponents. It's a power 4 question ... (2a3 - 5b2)4 If the variables were x and y, the answer would look like this: (we'll leave out the powers of 0) (x + y)4 = 1x4 + 4x3y1 + 6x2y2 + 4x1y3 + 1y4 Now replace x and y by (2a3) and (-5b2) = 1(2a3)4 + 4(2a3)3(-5b2)1 + 6(2a3)2(-5b2)2 + 4(2a3)1(-5b2)3 + 1(-5b2)4 and the rest is just easy algebra! Here is is: = 1(16a12) + 4(8a9)(-5b2) + 6(4a6)(25b4) + 4(2a3)(-125b6) + 1(625b8) = 16a12     - 160a9b2       + 600a6b4       - 1000a3b6        + 625b8 And there we have it ... a fast way to do a binomial expansion that uses 'choose' notation (or Pascal's Triangle), a simple observation about patterns, and some Grade 10 algebra. If you were doing these questions the long way, every increase in the power would mean many more lines and a lot more work. But using the method shown above, the same question with power 8 would involve just a little bit more algebra, but the same four lines of work! As with all new algebraic methods you are shown, you won't really learn it by watching it be explained. You need to practice it many times to get good at it. You learn best by 'doing'. Page 1 | Resources