SOLUTION: Find a polynomial with integer coefficients that satisfies the given conditions. Q has degree 3 and zeros −4 and 1 + i I do not know how to do this could you please hel

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Question 1099067: Find a polynomial with integer coefficients that satisfies the given conditions.
Q has degree 3 and zeros −4 and 1 + i

I do not know how to do this could you please help me.
Found 2 solutions by Boreal, Edwin McCravy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
the other zero is 1-i,since they are conjugate.
one factor is (x+4)
Using the quadratic formula, one wants to end up with [2+/- sqrt (-4)]/2, because that will reduce to 1+/-i
Therefore, using the quadratic formula, b=-2 and (b^2-4ac) must be -4, because the square root of -4 is +/-2i, and 2i/2=1, assuming a=1, which is the easiest case for Ax^2+Bx+C
That would be 4-4ac=-4 and a=1
so 4-4c=-4
-4c=-8
c=2
the quadratic factor is x^2-2x+2
The polynomial is the product of the two factors
x^3+4x^2-2x^2-8x+2x+8
x^3+2x^2-6x+8
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

Since 1 + i is a solution, so is its conjugate 1 - i

Set x = to each and get 0 on the right of each equation:

  x = -4;       x = 1+i;         x = 1-i
x+4 = 0;    x-1-i = 0;       x-1+i = 0

Multiply left sides and right sides of the three equations:

      (x+4)(x-1-i)(x-1+i) = (0)(0)(0)

Multiply and simplify:

  (x+4)[(x-1)-i][(x-1)+i] = 0  

         (x+4)[(x-1)²-i²] = 0

Square the binomial and replace i² by -1

      (x+4)[x²-2x+1-(-1)] = 0

         (x+4)[x²-2x+1+1] = 0

           (x+4)[x²-2x+2] = 0 

       x³-2x²+2x+4x²-8x+8 = 0

              x³+2x²-6x+8 = 0   

So the polynomial Q(x) is

                     Q(x) = x³+2x²-6x+8    

Edwin
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