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.
