SOLUTION: Find a polynomial equation with real coefficients that has the roots of -5, 2+i

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Question 1057047: Find a polynomial equation with real coefficients that has
the roots of -5, 2+i
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find a polynomial equation with real coefficients that has the roots of -5, 2+i
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The polynomial is p(x) = (x-(-5))*(x-(2+i))*(x-(2-i)) = (x+5)((x-2)-i)*((x-2)+i) = (x+5)*((x-2)^2+1) = (x+5)*(x^2 -4x + 5).

The polynomial equation is p(x) = 0.


Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find a polynomial equation with real coefficients that has
the roots of -5, 2+i
Since 2+i is a solution, and since the coefficients are all real,
its conjugate 2-i is also a solution.  Begin with

     x = 5;  x = 2+i;  x = 2-i

Get 0 alone on the right side of each:

   x+5 = 0;  x-2-i = 0; x-2+i = 0

Multiply all left sides together [and right sides too! (0)(0)(0)=0]

      (x+5)(x-2-i)(x-2+i) = 0

Simplify:

  (x+5)[(x-2)-i][(x-2)+i] = 0
         (x+5)[(x-2)²-i²] = 0
       (x+5)[(x-2)²-(-1)] = 0
          (x+5)[(x-2)²+1] = 0
      (x+5)[(x-2)(x-2)+1] = 0
         (x+5)[x²-4x+4+1] = 0
           (x+5)(x²-4x+5) = 0
             x³+x²-15x+25 = 0
  
So those zeros would be obtained when we set the
function f(x) = x³+x²-15x+25 equal to 0.  So
one answer is: 

f(x) = x³+x²-15x-25

Other answers would be gotten by multiplying the
right side by different constants k:
f(x) = kx³+kx²+25kx-15k

Edwin