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Mathematics
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this is a pain hold on
exactly haha
\[t(-1)=-4\] for a start, so you want \[\frac{t(x-1)+4}{x}\]
i wouldn't bother with the denominator under the 4? as a general question?
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\[\frac{4}{3(x-1)^3+2}+4\] is the next step
i am just saying that as a number, \[t(-1)=\frac{4}{3(-1)^2+2}=\frac{4}{-1}=-4\] so the numerator turns in to \[\frac{4}{3(x-1)^3+2}+4\] we can divide by x last
when you multiply this out and add these up you get \[\frac{12 x (x^2-3 x+3)}{3 x^3-9 x^2+9 x-1}\]
then divide by x to get \[\frac{12 (x^2-3 x+3)}{3 x^3-9 x^2+9 x-1}\]
who thought this up?????
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its online math hw for my college pre-calc class haha.
glad it is you and not me. enjoy
thanks though haha, the answers worked. Thanks Satellite
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