I am sure there is someone here that can easily tackle this problem! Factor: 3x^(-1/2)+9x^91/2)-81^(-3/2) I panic when I see negative fractional exponents! And my factoring skills are weak at best If you can explain any of the steps or the way you think about the problem I would greatly appreciate the advice.
\[ 3x^{\frac{-1}{2}}+9x^{\frac{1}{2}}-81^{\frac{-3}{2}}\] like this?
yes
I was thinking that an x^(1/2) If 81x^(-3/2) could be rewritten as 81x(-3(1/2)) and 3x(-1/2) could be rewritten as 3x^(-1(1/2))
i can write out what this means without negative exponents, but i cannot see how to factor it
oops that an x^(1/2) could be factored out *
oh wait, is there an x in the last term?
yes there was
\[3x^{\frac{-1}{2}}+9x^{\frac{1}{2}}-81x^{\frac{-3}{2}}\]
oh yikes I overlooked that
good catch. I looked over it so quickly the first time I missed the omission.
ok that is a whole different story. first off you can factor a 3 out of each term, then also perhaps an \[x^{\frac{1}{2}}\] if you like. lets try it
Great I get coefficients of 1 3 and 27
\[3x^{\frac{1}{2}}(x^{-1}+3+27x^{-2})\]
now i wonder if we can go further...
cool that exactly where I am at the moment as well
oh I had -27^(-3)
no i think it is \[27x^{-2}\]
because \[\frac{-3}{2}-\frac{1}{2}=-2\]
i cannot see how to factor the second part, so i think you are done at that step
ok so the exponents are not being multiplied in this case
I was thinking it was -27x^(-3(1/2))
sorry i was off by a minus sign, should be \[-27x^{-2}\]
when you multiply you add the exponents
so \[3x^{\frac{1}{2}}\times -27x^{-2}=-81x^{\frac{1}{2}+(-2)}=-81x^{-\frac{3}{2}}\] which is what you want
ok I think I see how that works so x^(1/2)x^-2(2/2) = x^(1/2)x^(-4/4) or x^(-3/2)
Thank you so much for your help! This really helped clarify the concept for me.
you don't need to change the denominator is 2
yw
oops meant (-2(2/2) = (-4/2)
got it
awesome!
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