find the excluded value pls...medal

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find the excluded value pls...medal

Mathematics
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(3x^2+5x-8)/(6x+16)*(4x^2+x)/(3x^2-3x)
2/(x^2-1)+x/(x^2-2x+1)

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let me rewrite that \(\Large \frac{(3x^2+5x-8)}{(6x+16)} \times \frac{(4x^2+x)}{(3x^2-3x)}\)
that's what you have?
for excluded values you look at what makes the denominator equal to zero
but before doing so we would like to simplify first factoring if possible
the answer was (3x^33-8)(4x+1)/6(3x+8)(x-1) @xapproachesinfinity
can we factor \(3x^2+5x-8\)
answer for what?
yes we can factor that @xapproachesinfinity
so what are the factors
(3x+8)(x-1) @xapproachesinfinity
yes!
so we have \(\Large \frac{(3x+8)(x-1)}{(6x+16)}\times \frac{(4x+1)}{3(x-1)}\) \(\Large \frac{(3x+8)(4x+1)}{3(6x+16)}\)
so what make denominator now equal to 0
6x+16=0 ===> x=-16/6 x=-8/3
excluded value -8/3
ok thanks can you help with the secondequation @xapproachesinfinity
actually i have to go for some time i fasting i need to go eat :)

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