Need help with math?

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Need help with math?

1. (x^3-8)/(x-2)

2.solve for z: y^2 +3yz-8z-4x

2007-09-01 08:06:34 · 3 answers · asked by jehu22293 2 in Science & Mathematics ➔ Mathematics

3 answers

(x^3-8) / ( x-2 ) is not defined when x = 2 because division by zero is undefined

(x^3 - 8 ) / ( x-2) = (x-2) (x^2+2x+4) / (x-2) = (x^2+2x+4)

y^2 + 3 yz - 8z -4x

2007-09-01 08:24:16 · answer #1 · answered by Will 4 · 0⤊ 0⤋

1. (x^3-8)/(x-2)=(x-2)(x^2+2x+4)/(x-2)=(x^2+2x+4)

y^2 +3yz-8z-4x= what

2007-09-01 15:44:45 · answer #2 · answered by Anonymous · 0⤊ 0⤋

1x^3-8=(x-2)(x^2+2x+4).
(x^3-8)/(x-2)=x^2+2x+4) AN.S.
2.y^2+3yz-8z-4x

=z(3y-8)= 4x-y^2
z=(4x-y^2)/3y-8)

2007-09-01 15:11:47 · answer #3 · answered by Anonymous · 0⤊ 0⤋