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(20064)  (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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