HELP!! how do you simplify ((x^3)-8)/(x-2)???

Question

also i need help solving....(5-X)/((x^2)-25)
please help ive been trying to figure these out for an hour

Answer 1

1- well for your first function you should do that
if f(x) = ((x^3)-8)) = [(x - 2)(x^2 + 2x + 4)]
so just put f(x) in fraction ;
(x - 2)(x^2 + 2x + 4) / (x-2)

now you can remove (x-2)
Therefor we have ;
x^2 + 2x + 4
now its another function so can find the roots .

2- f(x)= (5-X)/((x^2)-25)

first ((x^2)-25) = (5-X)/ (x+5)(x-5)
and (5-x) = (-x+5) ; facrot -1 ; so wed have - (x-5)
now put it in the fraction ;
f(x)= - (x-5) / (x+5)(x-5)
just remove (x-5)
Therefor we have
f(x) = -1 / (x+5)

Good Question.

Answer 2

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

so

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



about the 2nd

(x^2 - 25) = (x+5)(x-5)

so

(5-x)/(x^2-25) = -(x-5)/(x^2-25) = -1/(x+5)

Answer 3

(x^3-8)/(x-2)=
(x-2)(x^2+2x+4)/(x-2)
cancelling out x-2
the answer is x^2+2x+4

(5-x)/(x^2-25)=
-(x-5)/(x+5)(x-5)
cancelling out the x-5
the answer is -1/(x+5)

Answer 4

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

(5-x)/[x^2 - 25]
=(5-x)/[(x-5)(x+5)]
=[-(x-5)]/[(x-5)(x+5)]
= -1/(x+ 5)

HOpe that helps!

=)

Answer 5

(x^3 - 8)/(x - 2)
((x - 2)(x^2 + 2x + 4))/(x - 2)
ANS : x^2 + 2x + 4

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(5 - x)/(x^2 - 25)
(-x + 5)/(x^2 - 25)
(-(x - 5))/(x^2 - 25)
-(x - 5)/(x^2 - 25)
-(x - 5)/((x - 5)(x + 5))

ANS : -1/(x + 5)

Answer 6

Divide numerator and denominator by (x-2).
You get ( x^2 +x +2)

Answer 7

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