Dividing polynomials!! HELP! Urgent!!?

Question

Hey, I just got multiplying part of the polynomials, and now dividing is confusing me! I don't know how to go about this. Any ideas?

One

{(12m^2n^5)/(m+5)}/{(3m^3n)/(m^2-25)}

Two

{(n^2-n-12)/(2n^2-15n+18)}/{(3n^2-12n)/(2n^3-9n^2)}

Three

{(17mn^3)/(m^2+2m-35)}/{(34m^8n^4)/(m^2+7m)}

Four

{(16-2m)/(m^2+2m-24)}/{(m-8)/(3m+18)}

If anyone can help me on how to do these it will be great! I need advice and pointers, and then i can do the rest! There are 20 of these!!

I don't know why Y! Answers cuts the equations off, but here's the link with the questions:

http://oregonstate.edu/~patelu/scan0004.jpg

Help on any one of these will be great.

HELP

thank

Answer 1

One

{(12 m^2 n^5)/(m+5)}/{(3 m^3 n)/(m^2-25)}
= {(12 m^2 n^5)/(m+5)}*{(m^2-25)/(3 m^3 n) }
= {(12m^2n^5) *(m^2-25) } / { (m+5) * (3m^3n) }
= {(12 m^2 n^5) *(m+5)*(m-5)}/{(m+5)*(3 m^3 n) }
= {(12 m^2 n^5) *(m-5)}/(3 m^3 n)
= {(4 m^2 n^5) *(m-5)}/(m^3 n)
= {(4 m^2 n^4) *(m-5)}/m^3
= {(4 n^4) *(m-5)}/m
(Answer U)
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Three

{(17*m*n^3)/(m^2+2*m-35)}/
{(34*m^8*n^4)/(m^2+7*m)}
={(17*m*n^3)/(m^2+2*m-35)}*
{(m^2+7*m)/(34*m^8*n^4)}
={(17*m*n^3)*(m^2+7*m)}/
{(m^2+2*m-35)*(34*m^8*n^4)}
={(m*n^3)*(m^2+7*m)}/{ (m^2+2*m-35) *(2*m^8*n^4)}
={(n^3)*(m^2+7*m)}/
{(m^2+2*m-35) *(2*m^7*n^4)}
={(n^3)*m*(m+7)}/{(m^2+2*m-35) *(2*m^7*n^4)}
={(n^3)*(m+7)}/{(m^2+2*m-35) *(2*m^6*n^4)}
=(m+7)/{(m^2+2*m-35) *(2*m^6*n)}
=(m+7)/{(m+7)*(m-5)*(2*m^6*n)}
= 1/{(m-5)*(2*m^6*n)}
= 1/ {2*m^6*n*(m-5)}
(Answer L)
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Answer 2

long branch of polynomials is amazingly such as long branch of numbers. it works precisely like long branch of numbers could if digits ought to pass from minus infinity to infinity. With numbers you're able to commence on the leftmost digit - the optimal ability of 10 - and paintings your thank you to the appropriate. With polynomials, commence on the optimal ability of the variable and paintings your thank you to the appropriate. With numbers you're able to divide this leftmost piece of the dividend by ability of the divisor to fill interior the 1st digit of the quotient. With polynomials, that's analagous yet even less demanding. To fill interior the 1st term of the quotient, you in basic terms choose the optimal ability interior the divisor. what proportion cases does 3t^2 pass into 6t^4?. As with numbers, multiply out the divisor by ability of the recent piece of the quotient, and subtract. in case you have finished it authentic, the leftmost digit/term of the dividend will disappear. (3t^2+2t-4) * 2t^2 = 6t^4 + 4t^3 - 8t^2 (6t^4-5t^3-8t^2+15t-8) - (6t^4 + 4t^3 - 8t^2) = -9t^3 + 15t - 8 take care of something because of the fact the recent dividend and repeat the technique, till you attain the top of the quotient.