Factorization problems? T.T?

Question

x^2+2x(y+z)+(y+z)^2=
x^2-2x(y+z)+(y+z)^2=
x^2+6x(a+b)+9(a+b)^2=
x^2-6x(a+b)+9(a+b)^2=
3x(x-2y)^2-(x-2y)^3=
x^3(2x-y)+xy(2x-y)^2=
x^3(2x-y)-xy(2x-y)^2=
(a-b)(x^2-5)-(a-b)(3x+5)=

someone help T__T;
& can someone explain just a few of them so i get the idea?

Answer 1

The first few have to do with factoring, such as

x^2 + 4x + 3

In a case like this, you need to break it down into something like (x + ??)(x + ??) where the ?? are factors of the last term. For this, the last term is 3, and the only factors of 3 are 3 and 1. If we plugged those in, we'd have (x + 3)(x + 1), and if we used FOIL to expand that, we end up with x^2 + 4x + 3.

Sometimes you have to guess at the factors - if you were to factor x^2 - 5x + 6, factors of 6 are (6, 1), (-6, -1), (3, 2), and (-3, -2). In this case, you'd have to go through some trial and error to find that the factorization ends up being (x - 3)(x - 2).

Your first one is similar to my examples, except that instead of a 3 or a 6, your last term is (y + z)^2. But the idea is the same - find factors. In this case, the only factors are (y + z) and (y + z), so your factorization becomes

x^2 + 2x(y + z) + (y + z)^2 = [x + (y + z)][x + (y + z)]
= (x + y + z)^2, after you drop the parentheses and see that it's just the same thing, squared. It's always a good idea to use FOIL to check your work, too.

Your first four are going to be like this one - find the factors of the last term and use FOIL to be sure you piched them correctly.

Your last few are going to depend on you finding common factors in your terms. For example,

3x(x - 2y)^2 - (x - 2y)^3

...has a (x - 2y) term in common. So here, you'd factor this as

3x(x - 2y)^2 - (x - 2y)^3 = [(x - 2y)^2][3x - (x - 2y)]

You can keep going with this one:

= [(x - 2y)^2](2x - 2y)
= 2[(x - 2y)^2](x - y)

For these, just look for the common factrs and take them out. Once they're out, you can combine like terms like I did above and possibly factor some more.