SOLUTION: Please help factor the expression 1) y^2-16 2) n^4-m^4 3) 12a^4b^3c^4-108a^2bc^6 4) p^3-q^3=(p-q)

Question 217307: Please help factor the expression
1) y^2-16
2) n^4-m^4
3) 12a^4b^3c^4-108a^2bc^6
4) p^3-q^3=(p-q)
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
There are a variety of techniques that can be used when factoring:

Greatest Common Factor (GCF): Always start with this!
Factor by patterns (2 or 3 terms):
- Difference of Squares:
- Sum of Cubes:
- Difference of Cubes:
- Perfect Trinomial Squares:
Trinomial factoring (3 terms of the form )
Factoring by grouping (4 or more terms)
- If the groups have a common factor, factor it out. If not, then try different groupings. (Remember not everything will factor so there may be no way to form groups which will work.)
Trial and error of of the possible rational roots

There is no predetermined order to these methods except that you start with GCF factoring. After you factor out the GCF (unless it is 1), you try any and all of the others, in any order, as many times as necessary until nothing else will factor. It takes time to learn and get good at all of this.

Now let's try this on your problems:
1)
GCF is 1 which we don't usually factor out
Next, we try to figure out which other factoring method to try. After you practice enough of these you will see that this expression is a "difference of squares" since . So this will factor according the the pattern the the "a" being "y" and the "b" being "4":

This will not factor any further so we are finished.

2)
GCF = 1
This is another difference of squares: . So it will factor with the "a" being "" and the "b" being "":

We are not finished! Always check to see if you can factor anything else before you stop. If we look at what we have it should be obvious that the second factor is also a difference of squares so it will factor further:

NOTE: There is no pattern for "sum of squares". will not factor according to any pattern (or by any other factoring method. Since none of the factors will factor further we are now finished.

3)
GCF: Factoring this out we get:

The second factor is another difference of squares since so it will factor with the "a" being "ab" and the "b" being "3c":

None of these will factor further so we are finished.

4)
I'm going to assume that the equals sign is a typo and that the problem should be:
GCF: 1
There are 4 terms which is too many terms for any of the patterns or for trinomial factoring. So we'll try factoring by grouping. We should notice that the first two terms is a difference of cubes and we have a pattern for that. We should also notice that the factored form of the first two terms will include the last two terms as one of the factors which means factoring by grouping will work. Let's see how.
First group the first two terms and factor them as a difference of cubes: .

We can see (p-q) in both groups. So all we need to do to proceed is factor out a 1 from the second group. (Although we don't usually factor out 1's there are times when we do and this is one of those times):

Now we have (p-q) as a common factor between the two groups. So we can factor it out giving:

Take some time to look at this to see if you can understand it. This step of factoring by grouping may the the hardest one to learn. If it is not clear

think of the Distributive Property in reverse
multiply it back out and see if you end up with the original expression.

Neither factor will factor further so we are finished.
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