SOLUTION: how to solve:
12c to power of 2 divided by 3c+6
and
4x to power of 3 divided by 28x to power of 4
Algebra.Com
Question 411545: how to solve:
12c to power of 2 divided by 3c+6
and
4x to power of 3 divided by 28x to power of 4
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
4x^3 / 28x^4 would be a straight reduction.
divide numerator and denominator by 4 to get:
x^3 / 7x^4
since x^4 is the same as x^1*x^4, you can rewrite this expression as
x^3 / (7*x*x^3)
The x^3 cancels out in numerator and denominator to leave you with:
1/(7x)
you can also look at 4x^3 / 28x^4 in the following way.
4x^3 / 28x^4 is the same as (4/28) * (x^3)/(x^4)
x^3 / x^4 = x^(3-4) = x^(-1) = 1/x
you wind up with (4/28) * (1/x)
(4/28) is the same as (1/7) so you wind up with:
(1/7) * (1/x) which is the same as 1/(7x)
-----------------------------------
The result of dividing 12c^2 by (3c+6) results in (4c-8) with a remainder of 48.
This means that (3c+6) * (4c-8) + 48 will result in 12c^2
(3c+6) * (4c-8) equals 12c^2 -24c + 24c - 48
combine like terms to get:
(3c+6) * (4c-8) equals 12c^2 - 48
Add 48 to get a result of 12c^2
You're back to the original expression so the division was good.
Showing you how to do the division is difficult.
If you don't already know how to do it, the following reference would be helpful for you.
algebra long division
The first page shows you how to use factoring to divide.
The second page shows you how to do the long division.
The third page adds to that.
If you check algebra.com lessons, you might be able to find something about long division for algebra in there as well.
This was just easier for me to find.
It's fairly clear.
If you have problems with it, let me know and I'll do what I can do help you.
One other thing.
The answer of
(4c-8) with a remainder of 48 can also be shown as:
(4c-8 + 48/(3c+6)
You would check this result by multiplying (4c-8 + 48/(3c+6)) * (3c + 6).
The result will be 12c^2.
The long division is not terribly difficult.
Showing you how to do it here is long and time consuming.
The reference shows you the steps that you would take if you do not already know how to do them.
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