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can someone help me with this Algebra Problem? +Rep
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can someone help me with this Algebra Problem? +RepPosted:
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im kinda shy and dont like to ask questions and i dont like getting bad grades because i dont understand something. i was hopeing someone on here could help me do this problem.
Factor each Polynomial.
r4(to the fourth power) - k4 (sorry idk how to make them exponents, but they are.
Factor each Polynomial.
r4(to the fourth power) - k4 (sorry idk how to make them exponents, but they are.
#2. Posted:
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r4-k4
do you mean like that?
Edit: Correct me if I'm wrong but I think it's
(r2 + k2) (r2 - k2)
I'm not too sure tho.
Last edited by StoopidMonkie ; edited 1 time in total
do you mean like that?
Edit: Correct me if I'm wrong but I think it's
(r2 + k2) (r2 - k2)
I'm not too sure tho.
Last edited by StoopidMonkie ; edited 1 time in total
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#3. Posted:
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If you gave us k and r we will be able to figure that one out for you.
EDIT: Didn't see that you had factor
Answer: (r2+k2)(r-k)(r+k)
Last edited by -Tim ; edited 2 times in total
EDIT: Didn't see that you had factor
Answer: (r2+k2)(r-k)(r+k)
Last edited by -Tim ; edited 2 times in total
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#4. Posted:
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Fluttershyyy wrote r4-k4yeah its r4 - k4 i just cant make the numbers small
do you mean like that?
Edit: Correct me if I'm wrong but I think it's
(r2+k2)(r2-k2)
I'm not too sure tho.
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Fluttershyyy wrote r4-k4
do you mean like that?
Edit: Correct me if I'm wrong but I think it's
(r2+k2)(r2-k2)
I'm not too sure tho.
the second set can be simplified making the answer
(r2+k2)(r-k)(r+k)
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plug it in like this :p itll make it easier -(k-r)(k+r)(ksquared+rsqaured)
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#7. Posted:
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daaking3 wroteFluttershyyy wrote r4-k4
do you mean like that?
Edit: Correct me if I'm wrong but I think it's
(r2+k2)(r2-k2)
I'm not too sure tho.
the second set can be simplified making the answer
(r2+k2)(r-k)(r+k)
Oh your right....
Well there's your answer.
(r2 + k2) (r - k) (r + k)
/thread
/thread
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#8. Posted:
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if no one can figure it out its alright. but thanks to those who tried or are trying.
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#9. Posted:
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Mathway(Dot)com does it all for you!!
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