Find the factors : 5ab(2a-b) + 2b(b-2a)
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Answer:
If you substitute a=b+1 into (a^2−a)(2a−1) you get (b^2+b)(2b+1). Hence b+1-a is a factor of the expression you want factorised. It took a little time but I found the other factor to be 2a^2+2ab+2b^2-a+b. The only useful rule I can think of that comes from this example is that if a cubic in a and b can be factorised at all, then you should expect one of the factors to be linear (because if both factors were quadratic you would expect 4th degree terms in the product). Hence you could try substituting a=Xb+Y and see if there is a choice of X and Y which makes the expression come out to zero. In the case of the given problem, when you have made the substitution a=Xb+Y, there are two choices of X (either X=1 or X=-1) which make the coefficient of b^3 equal to zero and for either of these choices Y=1 is the only choice that makes the b^2 coefficient equal to zero ;and this also makes the constant term zero. The b coefficient is zero for X=Y=1. Hence, for the given problem, this explains why b+1-a is a factor.
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