Asked by Anonymous
Subtract:
F(x) = x^2+7x-8/x^3-5
G(x) = -7x-5x^3/4square root of x
F(x) = x^2+7x-8/x^3-5
G(x) = -7x-5x^3/4square root of x
Answers
Answered by
Steve
looks pretty straightforward, if I read it right.
f(x)-g(x) = (x^2+7x-8)/(x^3-5) - (-7x-5x^3)/(4√x)
so, it looks like we need to use a common denominator, so that gives us
(x^2+7x-8)(4√x) + (5x^3+7x)(x^3-5)
-------------------------------------------------------------
(4√x)(x^3-5)
Now just expand all that out if you wish. That pesky √x does get in the way, though.
f(x)-g(x) = (x^2+7x-8)/(x^3-5) - (-7x-5x^3)/(4√x)
so, it looks like we need to use a common denominator, so that gives us
(x^2+7x-8)(4√x) + (5x^3+7x)(x^3-5)
-------------------------------------------------------------
(4√x)(x^3-5)
Now just expand all that out if you wish. That pesky √x does get in the way, though.
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