Hello, dayla,

1) To calculate the zeros use substitution: y = x². Now you have the equation:

y² - 8y + 15 = 0 ==> y = 3 or y = 5

Re-substitute: x² = 3 or x² = 5. Thus you get: x1 ≈ -1.732 or x2 ≈ 1.732 or x3 ≈ -2.236 or x4 ≈ 2.236

With the second problem it is much easier to do C) first:

f(x) = x³ + 3x² + 9x + 27 = x²(x+3) + 9(x+3) = (x+3)(x²+9)

A) A function of third degree has 3 zeros, at least one real zero.

B) You have to solve:

0 = x³ + 3x² + 9x + 27 = x²(x+3) + 9(x+3) = (x+3)(x²+9)

A product is zero if one (or more) factor(s) are zero:

x + 3 = 0 ==> x = -3

x² + 9 = 0 ==> x² = -9 ==> x = 3ior x = -3i