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What are the real zeros in f(x)=x^3+8

User Drew C
by
8.5k points

2 Answers

4 votes

Answer:

X = -2

Explanation:

X^3 + 8 = 0

X^3 = -8

X = -8 ^ (1/3)

X = -2


User Purnima
by
7.0k points
3 votes

Answer:

-2

Explanation:

f(x)=x^3+8

Set it equal to 0

0 = x^2 +8

Subtract 8 from each side

-8 = x^3+8-8

-8 = x^3

Take the cube root of each side

-8 ^ 1/3 = x^3 ^ 1/3

-2 = x

The only real root is -2

User Gringogordo
by
8.8k points

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