4
(b-c)(a+b)-ac
If you mean (a-b+c)^2, then... a^2 - ab + ac - ab + b^2 - bc + ac - bc + c^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc.
(a + b)(b + c)
0
(a + b)(b - 2c)
(a + b)(b - 2c)
4
(b-c)(a+b)-ac
If you mean (a-b+c)^2, then... a^2 - ab + ac - ab + b^2 - bc + ac - bc + c^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc.
This is an algebra question. All you can do is sum similar terms. You get 6ab-5bc+2ac (same as 2ca) +ab+6bc-10ac = 7ab-8ac+bc
You want: abc + ab Factor out the common terms which are "a" and "b" ab ( c + 1 )
(a + b)(b + c)
(a + 3)( b + 2)
x2y + axy + abx + a2b Factor by grouping. xy(x + a) + ab(x + a) (xy + ab)(x + a)
ab + 4 + a + 4b factors to (a + 4)(b + 1)
The GCF is a.