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Question 173725: (1-r^3)
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(1-r)^3
you just have to factor & simplify i guess..
g(x)= x^3+2x over x^2-4
you need to know the domain, and the numbers to get zero , like on the numerator?
h(y)=(y^3-8)(y+2)^-3
g(s)=4s^2+15s-4 over (2s-1)^2
x^2+y^2-z^2-2xy over x^2-y2+z^2-2xz
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (1-r^3) / (1-r)^3
[(1-r)(1+r+r^2)]/[1-r]^3
= (1+r+r^2) / (1-r)^2
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g(x)= (x^3+2x) / (x^2-4)
= [x(x^2+2)] / [(x-2)(x+2)]
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Domain: All Real Numbers except x = 2 or x = -2
Zeroes: x = 0
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you need to know the domain, and the numbers to get zero , like on the numerator?
h(y)=(y^3-8)(y+2)^-3
= (y^3-8)/(y+2)^3
= [(y-2)(y^2 + 2y + 4)] / [(y+2)]^3
Domain: All Real Numbers except x = -2
Zeroes: y = 2
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g(s)=4s^2+15s-4 over (2s-1)^2
= [4s^2 +16s - s - 4]/[2s-1]^2
= [(4s-1)(s+4)]/[2s-1]^2
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Domain: All Real Numbers except s = 1/2
Zeroes: s = 1/4 and s = -4
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x^2+y^2-z^2-2xy over x^2-y2+z^2-2xz
I'll leave this one to you.
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Cheers,
Stan H.
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