SOLUTION: Find a polynomial with real corfficients that has -2 and 2 + i as zeros

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Question 1003892: Find a polynomial with real corfficients that has -2 and 2 + i as zeros
Found 2 solutions by ankor@dixie-net.com, jim_thompson5910:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Find a polynomial with real corfficients that has -2 and 2 + i as zeros
We have two factors from these zeros
x = -2
(x + 2) = 0
and
x = 2 + i
x - 2 = i
square both sides
(x-2)^2 = i^2
FOIL (x-2)(x-2); i^2 = -1
x^2 - 4x + 4 = -1
x^2 - 4x + 4 + 1 = 0
(x^2 - 4x + 5) = 0
Multiply these two factors
(x+2)(x^2 - 4x + 5) = x^3 - 2x^2 - 3x + 10 is polynomial
:
:
looks like this

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
x = 2+i is a root, so x = 2-i is another root. Complex roots come in conjugate pairs

Let's focus on x = 2+i. Get everything to one side (so 0 is on the other side).

x = 2+i
x - 2 = i
(x-2)^2 = i^2
(x-2)^2 = -1
x^2-4x+4 = -1
x^2-4x+4+1 = -1+1
x^2-4x+5 = 0

So we see that if 2+i is a root of p(x), then (x^2-4x+5) is a factor of p(x).
The same can be said about 2-i.
The same basic steps are followed so I won't show them.

If x = -2 is a root of p(x), then x-(-2) = x+2 is a factor of p(x)

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So we know that (x+2) and (x^2-4x+5) are factors of p(x)

Let's multiply the two factors

(x+2)(x^2-4x+5) = x(x^2-4x+5)+2(x^2-4x+5)
(x+2)(x^2-4x+5) = x*x^2+x*(-4x)+x*5+2*x^2+2*(-4x)+2*5
(x+2)(x^2-4x+5) = x^3-4x^2+5x+2x^2-8x+10
(x+2)(x^2-4x+5) = x^3-2x^2-3x+10

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So in the end, the polynomial is x^3-2x^2-3x+10 which is the final answer