SOLUTION: Form a polynomial with real coefficients having degree 3 and zeros -5 and 1+i

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Question 595748: Form a polynomial with real coefficients having degree 3 and zeros -5 and 1+i
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
A polynomial of degree with real coefficients will have three zeros within the set of complex numbers. And if any of the zeros are complex numbers then they will come in conjugate pairs. So if 1 + i is a zero then 1 - i will also be a zero. This makes the three zeros -5, 1+i and 1-i and the polynomial will have the following factors:
(x - (-5))(x - (1 + i))(x - (1 - i))
Simplifying we get:
(x + 5)(x - 1 - i))(x - 1 + i))

To get a polynomial we just need to multiply this out. Multiplying this out, especially with the two trinomials, looks like it will be a lot of work. Here's a way to make this easier. First I'll regroup the last two factors:
(x + 5)((x - 1) - i))((x - 1) + i))
Looking at the last two factors in this way we can see that they fit the pattern:
%28a-b%29%28a-b%29+=+a%5E2+-+b%5E2
with "a" being (x-1) and "b" = i. We can use the pattern to multiply these factors quickly and easily:
%28x+%2B+5%29%28%28x+-+1%29%5E2+-+i%5E2%29%29
which simplifies as follows:
%28x+%2B+5%29%28%28x%5E2+-2x+%2B+1%29+-+%28-1%29%29
%28x+%2B+5%29%28x%5E2+-2x+%2B+2%29
Multiplying the remaining factors:
x%5E3+-+2x%5E2+%2B+2x+%2B5x%5E2+-10x+%2B10
Simplifying:
x%5E3+%2B+3x%5E2+-8x+%2B10
This is a polynomial of degree 3 with the given zeros.

Note: Several references were made to "a polynomial", not "the polynomial". This is due to the fact that any multiple of this polynomial will also fit the description:
2x%5E3+%2B+6x%5E2+-16x+%2B20
10x%5E3+%2B+30x%5E2+-80x+%2B100
-x%5E3+-+3x%5E2+%2B8x+-10
etc.