Respuesta :

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1) Find the real and complex solutions of x^3 - 216 = 0

Answer: fourth choice 6, - 3+3√3 i, and -3 - 3√3 i

Solution:

1) Factor x^3 - 216 as a difference of cubes:

(x)^3 - (6)^3 = (x - 6) (x^2 + 6x + 6^2) = (x - 6) (x^2 + 6x + 36) = 0

3) first factor = 0 => x - 6 = 0 => x = 6

4) second factor = 0 => x^2 + 6x + 36 = 0

Using the quadratic formula:

x = [ - 6 +/- √(6^2 - 4*1*36) ] / (2)

=> x = [ - 6 +/- √ (-108) ] / 2 = -3 +/- 3√3 i

=> x = - 3 + 3√3 and x = - 3 - 3√3 i

Answer: 3, - 3+3√3 i, and -3 - 3√3 i

2) Which is equivalent to (2p^2 + 5pq - q^2) + (p^2 + 3pq - 2q^2) ?

Answer: first option 3p^2 + 8pq - 3q^2

Solution:

1) eliminate parenthesis:

2p^2 + 5pq - q^2 + p^2 + 3pq - 2q^2

2) combine like terms:

2p^2 + p^2 = 3p^2

5pq + 3pq = 8pq

-q^2 - 2q^2 = - 3q^2

3) The result is 3p^2 + 8pq - 3q^2, which is the first option.



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