ANSWERS: 2

(x^2  1)(x  1)^2 = (x^2  1)(x^2  2x + 1) = x^4  2x^3 + x^2  x^2 + 2x  1 = x^4  2x^3 + 2x  1 First off I chose (x  1)(x + 1)^2 because I needed a 1 constant. I knew that I'd be miising a multiple of x so with a little trial and error I got the above.

Farino may be right, I havent checked his work. But right off the bat what I dont like is that he is starting with the solution and then proving it works. I dont think that is very appropriate an explanation when someone asks how to factor. Here is what I would do... x^4  2x^3 + 2x  1 Factor 2x out of the second and third terms: x^4  2x(x^2  1)  1 The first and fourth terms are a difference of squares: x^4  1  2x(x^2  1) (x^2  1)(x^2 + 1)  2x(x^2  1) Now you have (x^2  1) as a liketerm (x^2  1)[ (x^2 + 1)  2x ] Simplify (x^2  1)(x^2  2x + 1) The right term is a quadratic that can be factored (x^2  1)(x  1)^2 ...... this is what Farino has. The left is a difference of squares (x  1)(x + 1)(x  1)^2 Simplified (x + 1)(x  1)^3 Now... unfortunately there is no explicit procedure. There are a million paths you can take. And sometimes you take the wrong one and have to backtrack. It just takes algebraic manipulation... and a bit of intuition for the patterns. Sometimes you may have to create new arbitrary terms and stick them in, add and subtract the same value to balance the equation, just to make an algebraic rule work.
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