Since x^2 = 16a^2(b-4)^2, the quantity you want to factor is just:x^2 - xx(x-1)Now you can plug x back in:4a(b-4) * [(4a(b-4)) - 1]Again, you can expand that stuff on the right to get your answer.-3x(x+2)+ 5(x+2) -->each term has one same factor right? and that is x+2, all you have to do is combine the two different factors (3x+5) then copy the same factor (x+2) so the answer will be (3x+5)(x+2)16a^2(b-4)^2 - 4a(b-4) -->okay,......
I can see how this gets confusing now, because there are two strange situations. The first strange situation is that the second term COMPLETELY divides the first term. The second strange situation is that the first term is the SQUARE of the second term.
One technique that may make factoring stuff like this easier is using substitution. Make some variables for stuff that appears often. For example, 4a(b-4) appears often. So, you might want to let x = 4a(b-4). Since x^2 = 16a^2(b-4)^2, the quantity you want to factor is just:
x^2 - x
x(x-1)
Now you can plug x back in:
4a(b-4) * [(4a(b-4)) - 1]
Again, you can expand that stuff on the right to get your answer.
3x(x+2)+ 5(x+2) -->each term has one same factor right? and that is x+2, all you have to do is combine the two different factors (3x+5) then copy the same factor (x+2) so the answer will be (3x+5)(x+2)
16a^2(b-4)^2 - 4a(b-4) -->okay, this also like what i explain above, let's do it again
[16a²(b-4)-4a](b-4) -->b-4 on both sides are same but one is squared, so copy b-4, then add the remaining factors including one b-4[remember that (b-4)² is also (b-4)(b-4)]
[16a²(b-4)-4a](b-4) -->distribute first the factor on first term
(16a²b-64a²-4a)(b-4) -->simplify the first factor since all terms are divisible by 4
4a(4ab-16a-1)(b-4) --> final answer, it's also the same as 4a(b-4) (4ab-16a-1)
