Question 629144: form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree; zeros:-1;-i;-5+1 Answer by jsmallt9(3758) (Show Source):
So all I can do right now is give you some pointers:
The degree will tell you how many zeros f(x) will have.
f(x) can be expressed as the product of factors of the form:
(x - z)
where the "z" is one of the zeros. Since the degree tells you the number of zeros, it also tells us how many of these factors there will be.
When a polynomial with real coefficients has a complex zero, the complex conjugate of that zero will also be a zero. For the standard form complex number:
a + bi
its complex conjugate is
a - bi (or a + (-bi))
Your zeros include -i which is a complex number. In standard form -i is:
0 + (-i)
This makes its complex conjugate:
0 - (-i) or 0 + i or just i
So your polynomial will have -i and i as zeros which means that (x-i) and (x-(-i)) (or (x+i)) will be factors of f(x).
Putting this together into a procedure:
Determine the degree of f(x)
Find all the zeros of f(x). This may require finding complex conjugates to "fill in" for zeros you were not given.
Write f(x) as the product of factors where each factor is of the form (x - z) and "z" is one of the zeros. Of the zeros we know from what you posted you would have:
f(x) = (x-(-1))(x-(-i))(x-i)...
which simplifies to:
f(x) = (x+1)(x+i)(x-i)...
Multiply all the factors. You probably know that the order does not matter when you are multiplying. And that is the case here, too. But you can make things easier on yourself if you multiply the factors with the complex conjugate zeros first. So start with (x+i)(x-i) first. Then multiply any other pairs of factors with complex conjugate zeros. Then multiply the rest. By using this order
You can use the pattern to multiply quickly; and
The i's will all disappear quickly making the remaining multiplications easier.
If this is not enough for you to figure out how to do the problem, then please re-post the correct and complete problem.
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