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March 27, 2015

March 27, 2015

Posted by **-Untamed-** on Saturday, December 31, 2011 at 7:51pm.

(3t^2+3t-6/2t^2-2t-4)(4t^2+4t-24/3t^2+6t-9)

This is the way I tried solving this:

3t(t-1)6(t-1)/2t(t-1)4(t-1)*4t(t-2)12(t-2)/3t(t+1)9(t-1)

(3)(t)(t-1)(2)(3)(t-1)/(2)(t)(t-1)(2)(2)(t-1)*(2)(2)(t)(t-2)(2)(2)(3)(t-2)/(3)(t)(t-1)(3)(3)(t-1)

After cancelling out all the terms I am left with:

(2)(t-2)(t-2)/(t-1)(t-1)

^ I cant cancel anything off now. But the answer is supposed to be :

2(t+2)/t+1

^ How do I get that answer?

- Calculus -
**Damon**, Saturday, December 31, 2011 at 8:12pm(3t^2+3t-6)(4t^2+4t-24)

===============================

(2t^2-2t-4)(3t^2+6t-9)

3(t^2+t-2)(4)(t^2+t-6)

==============================

2(t-2)(t+1)(3)(t+3)(t-1)

3(t+2)(t-1)(4)(t+3)(t-2)

==============================

2(t-2)(t+1)(3)(t+3)(t-1)

(2) (t+2) / (t+1)

2

- Calculus -
**-Untamed-**, Saturday, December 31, 2011 at 8:17pmThank you so much Damon :)

- Calculus -
**Damon**, Saturday, December 31, 2011 at 8:21pmYou are welcome.

By the way that is undefined if t = -1 so I suppose that might be what is meant by "non-permissible"

- Calculus -
**-Untamed-**, Saturday, December 31, 2011 at 8:25pmThank you and yeah nonpermissible is another word for restrictions.

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