Please a little help with 2 extra credit factoring problems

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obstrep

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Oct 12, 2010
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Today in class I was given an 2 extra credit assignments that I cannot figure out. I know this will be a huge problem I have tried to work it out but I really am not sure if it is correct. I would greatly appreciate any help cause I could really use the little extra credit towards my grade. I need it done by sometime tomorrow 10/13/10. Thanks in advance :)

(A+B+C+D) to the fourth power


then I have

A to the 6th power minus B to the 6th power
 
Re: Please a little help with 2 extra credit factoring probl

obstrep said:
Today in class I was given an 2 extra credit assignments that I cannot figure out. I know this will be a huge problem I have tried to work it out but I really am not sure if it is correct. I would greatly appreciate any help cause I could really use the little extra credit towards my grade. I need it done by sometime tomorrow 10/13/10. Thanks in advance :)

(A+B+C+D) to the fourth power


then I have

A to the 6th power minus B to the 6th power
Click to expand...

You say you have tried to work it out. What have you done? You say "I really am not sure if it is correct." We aren't sure either, because you have shown us nothing.

You say "I need it done by sometime tomorrow 10/13/10".....then I guess you had better get to work on it.

If you want HELP, we need to see what you've done.

If you want someone to do it for you, so that YOU can get the extra credit, you've come to the wrong place.
 
Re: Please a little help with 2 extra credit factoring probl

Sorry I didn't post my work fisrt. I read the first posting and it said not to post work you have already done.Since I'm new here I was trying to follow the rules.
for a to the 6th minus b to the 6th
(a^3-b^3) (a^3+b^3)

(a-b)(a^2+ab+b^2)(a-b)(a^2+ab+b^2)

(a-b)^2 (a^2+ab+b^2)^2


I'm still working on the longer one but I think Im just confusing myself. Im not looking for anyone to anwser my problems for me I could go to the math tutor at school for that. I want help is all.
 
Re: Please a little help with 2 extra credit factoring probl

obstrep said:
Sorry I didn't post my work fisrt. I read the first posting and it said not to post work you have already done.Since I'm new here I was trying to follow the rules.
for a to the 6th minus b to the 6th
(a^3-b^3) (a^3+b^3)

(a-b)(a^2+ab+b^2)(a-b)(a^2+ab+b^2)

(a-b)^2 (a^2+ab+b^2)^2


I'm still working on the longer one but I think Im just confusing myself. Im not looking for anyone to anwser my problems for me I could go to the math tutor at school for that. I want help is all.
Click to expand...

I'm quite sure that our "rules for posting" ask that you DO show your work.

Now...your first step on factoring

a[sup:2fnivh9d]6[/sup:2fnivh9d] - b[sup:2fnivh9d]6[/sup:2fnivh9d]
is appropriate...you factored as the difference of two squares:

(a[sup:2fnivh9d]3[/sup:2fnivh9d] - b[sup:2fnivh9d]3[/sup:2fnivh9d])(a[sup:2fnivh9d]3[/sup:2fnivh9d] + b[sup:2fnivh9d]3[/sup:2fnivh9d])

And you've correctly attempted to factor each of those binomials...but made a small error.

(a[sup:2fnivh9d]3[/sup:2fnivh9d] - b[sup:2fnivh9d]3[/sup:2fnivh9d]) is a difference of two cubes....and you factored that correctly as
(a - b)(a[sup:2fnivh9d]2[/sup:2fnivh9d] + ab + b[sup:2fnivh9d]2[/sup:2fnivh9d])

(a[sup:2fnivh9d]3[/sup:2fnivh9d] + b[sup:2fnivh9d]3[/sup:2fnivh9d]) is a SUM of two cubes. You may wish to review the pattern for factoring such a thing...you've factored it the same way as the difference of two cubes.
 
Re: Please a little help with 2 extra credit factoring probl

obstrep said:
Today in class I was given an 2 extra credit assignments that I cannot figure out. I know this will be a huge problem I have tried to work it out but I really am not sure if it is correct. I would greatly appreciate any help cause I could really use the little extra credit towards my grade. I need it done by sometime tomorrow 10/13/10. Thanks in advance :)

(A+B+C+D) to the fourth power
____________________________________________________________________
Are you sure they wanted above to be factorized?

The only way that can be approached is:

\(\displaystyle a^4 \ = \ a \ * \ a \ * \ a \ * \ a\)

Or did they want the above to be "de-factorized" (is that even a word?) like:

\(\displaystyle (a+b)^2 \ = \ a^2 \ + \ 2ab \ + \ b^2\)
_____________________________________________________________________
then I have

A to the 6th power minus B to the 6th power
Click to expand...
 
Re: Please a little help with 2 extra credit factoring probl

Thanks for the help
I think I figured out what you said.
Would the correct anwser be
(a-b)(a+b)(a^2-ab+b^2)^2
 
Re: Please a little help with 2 extra credit factoring probl

obstrep said:
Thanks for the help
I think I figured out what you said.
Would the correct anwser be
(a-b)(a+b)(a^2-ab+b^2)^2 .............. no
Click to expand...

\(\displaystyle a^3 - b^3 \ = \ (a-b)(a^2 \ + \ ab+b^2)\)

\(\displaystyle a^3 + b^3 \ = \ (a+b)(a^2 \ - \ ab+b^2)\)

Watch those signs carefully.....
 
Re: Please a little help with 2 extra credit factoring probl

It's solved so far the correct way factoring the sum or difference of two cubes. I have some simple examples in my book that are set up in this format but this one was much harder. thanks
 
Re: Please a little help with 2 extra credit factoring probl

Ok here's my work to the other problem (a+b+c+d)^4
Please let me know if I messed up becaus ethis was NOT easy.
(a+B+c+d)^4
[(a+b)+(c+d)]^4
(a+b)^2+(c+d)^2+2(a+b)(c+d)^2
(a^2+2ab+b^2+c^2+d^2+2cd+(2a+2b)(c+d)^2
(a^2+b^2+c^2+d^2+2ab+2cd+d^2+2ac+2ad+2bc+2bd)^2
(a^2+b^2+c^2+d^2+2ab+2bc+2ca+2cd+2bd)^2[(a+b)^2+(c+d)^2+2(a+b)(c+d)]^2
(a+b)^4+(c+d)^4+[2(a+b)(c+d)]^2+2(a+b)^2(c+d)^2+4(c+d)^3(a+b)+2(a+b)^3(c+d)

That's a lot of numbers and letters
 
Re: Please a little help with 2 extra credit factoring probl

obstrep said:
Ok here's my work to the other problem (a+b+c+d)^4
Please let me know if I messed up becaus ethis was NOT easy.
(a+B+c+d)^4
[(a+b)+(c+d)]^4
(a+b)^2+(c+d)^2+2(a+b)(c+d)^2
Click to expand...
I think this is about the most ridiculous exercise I've seen ! :shock:

Anyway, you should at least CHECK if you're correct at each step;
let a = b = c = d = 1
(a+b+c+d)^4 = 4^4 = 256

Then your (a+b)^2+(c+d)^2+2(a+b)(c+d)^2 should also equal 256:
2^2 + 2^2 + 2(2)(2^2) = 4 + 4 + 16 = 24 : so WHOA!!

Get my drift?

Btw, my sacarsm is directed at your teacher, not at you... :wink:
 
Re: Please a little help with 2 extra credit factoring probl

\(\displaystyle [a + b + c + d]^4 \ = \ \left [[(a + b) + (c + d)]^2\right ]^2 \ = \ \left [[(a + b)^2 + (c + d)^2 + 2(a+b)(c+d)]\right ]^2\)

And continue .... long way to go ..... your teacher does not like you .....
 
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