Given that f(x) is a cubic function with zeros at -7 and -2i - 2, find an equation for f(x) given that f(0)=6.

Answer 1

If one of the zeros is -2i - 2 (although let's write it properly as -2 - 2i), then by the Conjugate Pair Theorem, -2 + 2i is also a zero. so now we have 3 zeros, -7, -2 - 2i, and -2 + 2i. Rewrite those as factors by prefixing each with "x -" to get [x - (-7)] or (x + 7), [x - (-2 - 2i)] or (x + 2 + 2i), and [x - (-2 + 2i)] or (x + 2 - 2i). Multiply these together. First multiply (x + 2 + 2i) and (x + 2 - 2i) to get x² + 4x + 8 then multiply that by (x + 7) to get x³ + 11x² + 36x + 56 however, this could be multiplied by a factor and still have the same zeros, so we must put a facor in front of it. Let's call the factor "A". That means the polynomial, so far, is y = A(x³ + 11x² + 36x + 56). Now, let's use the fact that f(0) = 6. Plug those x and y values in to get 6 = A((0)³ + 11(0)² +36(0) + 56) or A = 3/28. The final polynomial is f(x) = 3/28(x³ + 11x² + 36x + 56) or f(x) = 3/28x³ + 33/28x² +27/7x + 6

Answer 2

We first define a function g(x) which will be a cubic function with zeros at x = -7 and x = -2 - 2i. Before doing this, recall that if a complex number a - bi is a zero, then so is its conjugate a + bi. So, we have a third zero at x = -2 +2i. We proceed now as follows:

x = -7 <--> x + 7 = 0

x = -2 - 2i <--> x + (2 + 2i) = 0

x = x = -2 + 2i <--> x + (2 - 2i) = 0

We therefore define g(x) = (x + 7) (x + (2 + 2i)) (x + (2 - 2i)). Focus now on

(x + (2 + 2i)) (x + (2 - 2i)) = x² + (2 + 2i)x + (2 - 2i)x + (2 + 2i)(2 - 2i)

= x² + 4x + 4 - 4i²

= x² + 4x + 8

We can update

g(x) = (x + 7) (x² + 4x + 8)

We want to find f(x) so that f(0) = 6. Notice g(0) = (0 + 7)(0 + 0 + 8) = 56. With this in mind, let

f(x) = ⁶/₅₆ g(x) , because f(0) = ⁶/₅₆ g(0) = ⁶/₅₆ (56) = 6

= ³/₂₈ (x + 7) (x² + 4x + 8)

= ³/₂₈ (x³ + 4x² + 8x + 7x² + 28x + 56)

= ³/₂₈ x³ + ³³/₂₈ x² + ²⁷/₇ x + 6