SOLUTION: Create a degree 3 polynomial in the form 𝑓(𝑥) = 𝑎x^3 + 𝑏𝑥^2 + 𝑐x + 𝑑 such as f(6) = 0 and f(-2 - 3i) = 0.

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Question 1089702: Create a degree 3 polynomial in the form 𝑓(𝑥) = 𝑎x^3 + 𝑏𝑥^2 + 𝑐x + 𝑑
such as f(6) = 0 and f(-2 - 3i) = 0.
Found 2 solutions by jim_thompson5910, ikleyn:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

f(6) = 0 means that x = 6 is one root
x = 6
x - 6 = 0
So x-6 is one factor of the cubic polynomial
----------------------
f(-2-3i) = 0 means x = -2-3i is another root
x = -2-3i
x+2 = -3i
(x+2)^2 = (-3i)^2
(x+2)^2 = 9i^2
(x+2)^2 = 9(-1)
(x+2)^2 = -9
x^2+4x+4 = -9
x^2+4x+4+9 = 0
x^2+4x+13 = 0
and x^2+4x+13 is the other factor of the cubic polynomial
----------------------
Multiply out the two factors x-6 and x^2+4x+13
(x-6)*(x^2+4x+13)
x*(x^2+4x+13)-6*(x^2+4x+13)
x^3+4x^2+13x-6x^2-24x-78
x^3-2x^2-11x-78
-------------------------------------------------------
-------------------------------------------------------
Answer: f(x) = x^3-2x^2-11x-78
Note how
a = 1
b = -2
c = -11
d = -78
It's possible to scale this polynomial so it's technically not the only possible answer; however, this is the form where the GCF of the four coefficients is 1. So in a sense, this is the most reduced version.
Answer by ikleyn(52909)   (Show Source): You can put this solution on YOUR website!
.
Notice that your question was mathematically incorrect.

The correct version is THIS: 

    Create a degree 3 polynomial in the form 𝑓(𝑥) = 𝑎x^3 + 𝑏𝑥^2 + 𝑐x + 𝑑
     such as f(6) = 0 and f(-2 - 3i) = 0.


Only if the coefficients of a polynomial are real, then the complex roots go as complexity-conjugated pairs.


Which is NOT TRUE for the general polynomial with complex coefficients.



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