Expert: Paul Klarreich - 1/8/2006

Question
How are you well, i need to factor out these problems completly, and it is for alegrbra 2. thank you. i can also fax over the work sheets to you, cause i am taking trig and alg 2 and it is hard and i need help.
Question -
1) x^6+125y^6
2) 8c^3-64d^3
3) y^3-27
4) a^3+1
5) y^8-38y^4+72
6) c^2^x-49D^2
7) 16a^2-4b^2

^=stands for exponents


Answer
Hi, Val,

Your question - Factor these completely.

{That's better -- now I know what you need.)
You wrote:
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How are you well, i need to factor out these problems completly, and it is for algebra 2. thank you. i can also fax over the work sheets to you, cause i am taking trig and alg 2 and it is hard and i need help.
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Well, I think that is going too far, and there isn't any good way to fax them.  Besides, I am not allowed to answer that many questions at a time.  So it is best that you pick out the ones you are having trouble with and send them along.  Perhaps the act of having to type them carefully into the computer may in fact give you the clue.
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Here is a general scheme for these.  You want to match the example to one these general factoring patterns:  (I am using p and q because they don't appear in your examples. Usually we write x and y or a and b.)

I.   p^2 - q^2 = (p + q)(p - q) -- a difference of squares.
IIA. p^3 + q^3 = (p + q)(p^2 - pq + q^2) -- a sum of cubes.
IIB. p^3 - q^3 = (p - q)(p^2 + pq + q^2) -- a difference of cubes.
Or a trinomial of the form:
III. ap^2 + bp + c  for which there are other methods.

So you decide, looking at it, whether it has 2 or 3 terms.  If it has two terms, it is like I or II; if it has three, it is like III.

1) x^6+125y^6

Two terms and not a difference (it has + in the middle).  But

x^6 = (x^2)^3,  a cube
125y^6 = (5y^2), a cube.
So this matches IIA with  p = x^2  and q = 5y^2 and we fill in the right side:

(p + q)(p^2 - pq + q^2)
(x^2 + 5y^2)(x^4 - 5x^2y^2 + 25y^4)

Hopefully this will be enough to set the pattern for the other examples and I can be briefer.(much)

2) 8c^3-64d^3  is a difference of cubes, IIB, p=2c, q=4d

HOWEVER, your instructions say Factor Completely, and that means to take out a common factor first:

8c^3-64d^3 = 8(c^3 - 8d^3), and the second factor is a IIB with p=c, q=2d

= 8(c - 2d)(c^2 + 2cd + 4d^2)


3) y^3-27 is a difference of cubes.  Use IIB.(I leave that to you.)
4) a^3+1  is a sum of cubes.  Use IIA. (I leave that to you.)

5) y^8-38y^4+72 can be made to match III, using p = y^4.  Then it looks like:

p^2 - 38p + 72 = (p - 36)(p - 2) = (y^2 - 36)(y^2 - 2)


6) c^2^x-49D^2   (ARE YOU SURE IT DIDN'T SAY  c^2x?)

This is tricky, because the exponents don't commute. That is to say, c^2^x and c^x^2 are not, in general, the same. Remember that repeated exponents like these are always evaluate from the top down.  For example:

c^2^5 = c^(2^5) = c^32

Evaluating from the bottom would give c^10 which is wrong.

But  c^2^x = [c^2^(x-1)]^2.  That's not obvious, but think of, perhaps,

c^2^5 = c^32 = (c^16)^2 = (c^2^4)^2


So your example works this way:

c^2^x-49D^2 is a difference of squares, p = c^2^(x-1),  q = 7D

= (c^2^(x-1) + 7D)( c^2^(x-1) - 7D)

7) 16a^2-4b^2  shouldn't be too hard now.  It is a difference of squares and you should be able to handle it.