SOLUTION: I am hoping I am placing this is the correct category as I did not see one for rational expressions and equations.
I know this seems fairly simple but for some reason I am extrr
Algebra.Com
Question 246820: I am hoping I am placing this is the correct category as I did not see one for rational expressions and equations.
I know this seems fairly simple but for some reason I am extrremly confused ....i would really appreciate some help and explantion as to what i might be doing wrong.....
1) x^5-4x^3, x^3+4x^2+4x Instructions state to find LCM
My answer: 2*2*x*x*x*x*x*x*x*x, 2*2*2*2*x*x*x*x*x*x
=2*2*2*2*x*x*x*x*x*x*x*x
answer: 16x^8
2)x^2/x-7 + 49/7-x
answer: 49x^2 (I believe x-7 and 7-x cancel out. I am very confused though and when looking at the other examples in the book and doing the other exercises it does not look like my answers are correct. I am unsure as to where exactly I am going wrong or what I am simply not getting. Thank you so much for any guidance you can give me.....I truly appreciate it.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1) x^5-4x^3, x^3+4x^2+4x Instructions state to find LCM
To find any LCM find the prime factors of the numbers you are given:
x^5-4x^3 = x^3(x^2-4) = x^3(x-2)(x+2)
----------------------------------------
x^3+4x^2+4x = x(x^2+4x+4) = x(x+2)^2
-----------------------------------------
Then the LCM is the product of each of the prime factors to their
highest power.
--------------------------
LCM = x^3*(x-2)(x+2)^2
=======================================
2)x^2/(x-7) + 49/(7-x)
---
Rewrite as:
x^2/(x-7) - 49/(x-7)
--------------------------
Since the denominators are the same, combine the numerators
over the lcm.
(x^2-49)/(x-7)
Factor:
[(x+7)(x-7)]/(x-7)
Cancel to get the final answer:
x+7
==============
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Start with 
:
Take out a 
, leaving you with )
, then factor the difference of two squares, leaving you with (x\ +\ 2))
Now look at 
:
Take out an 
, leaving you with )
and then factor the trinomial leaving you with (x\ +\ 2))
The least common multiple requires:
3 factors of (because the factor that has the most factors of has 3 -- one has three and the other has one)
2 factors of 
(because the factor that has the most factors of has 2 -- one has one and the other has two)
and
1 factor of 
(because the factor that has the most factors of has 1 -- one has one and the other has none)
Hence the LCM is:
^2(x\ -\ 2))
Once you multiply it out it will be a very handsome little 6th degree polynomial. Well, handsome being a relative term anyway. It will certainly be prettier than the 8th degree horror you would have had if you had simply multiplied by 
=====================================
Multiply 
by -1:
\ =\ (-1)\,\cdot\,7\ -\ (-1)\,\cdot\,x\ =\ -7\ -\ (-x)\ =\ -7\ +\ x\ =\ x\ -\ 7)
See? In general, )
The only specific case where 
is where 
.
So, for your problem:
Multiply either of the fractions by -1 over -1:
)
Now you have the denominators equal, so:
Now factor the difference of two squares numerator:
(x\ -\ 7)}{x\ -\ 7})
And then eliminate the common factor from the numerator and denominator by applying the multiplicative identity: 
. (Please eliminate the word "cancel" from your mathematical vocabulary. The misuse of that idea has screwed up more math students than could ever be counted)
But remember to specify that
because, while 14 is a perfectly good value for 
, taking on a value of 7 would cause you to have a zero denominator in both of your original fractions. If you are not yet comfortable with this idea, get a graphing calculator or graphing program for your computer, graph the function:
\ =\ \frac{x^2\ -\ 49}{x\ -\ 7})
And see what the calculator or computer tells you when you try to evaluate
)
[TILT!!]
John
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