I got an excercise which i got stuck, hope some one can help. It's about factoring by grouping.
This is the problem $ab^2 - 2b + 3a - 6$
This is what I tried... $$ab^2 - 2b + 3a - 6$$
$$(ab^2 - 2b)+(3a - 6)$$
$$b(ab -2) + 3(a-2)$$
I don't know how to continue from here since $(ab-2) ≠ (a-2)$.
In the documentation I have it only says that the answer to this problem is...
$$b^2(a-2) + 3(a-2)$$ $$(a-2)(b^2 + 3)$$
But it doesn't show the steps before this. Assuming my first steps are correct, what doesn't make sense to me is why he changed $b(ab -2)$ to $b^2(a-2)$ they don't produce the same terms
$$b^2(a-2) ≠ b(ab - 2)$$ $$ab^2 - 2b^2≠ ab^2 -2b$$
Are you sure the answer is right? I agree with you. Because, $ab²-2b+3a-6=a(b²-2)+3(a-2)$ or,
$ ab²-2b+3a-6=ab²+2b-4b-4-2+3a=a(b²+3)-2(-b+2b+2+1)=a(b²+3)-2(b-3) $.