Question 424510: find a fourth degree polynomial function with real coefficients satisfying the given conditions:
n=3; 3 and i are zeros; f(2)=25 Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website! A fourth degree polynominal with real coefficients will have 4 roots. You are given only three of them. To find the 4th root you have to know that the complex roots of such polynomials will always come in conjugate pairs. So we will be able to find the 4th root using the root, i. In full standard form for complex numbers,
i = 0 + i
The complex conjugate of 0 + i is 0 + (-i) or simply -i. So the 4 roots of the desired polynomial are 3, 3, i and -i.
With these roots we can find the polynomial. If some number "r" is a root of a polynomial, then (x-r) will be a factor of the polynomial. So our polynomial will have factors of (x-3), (x-3), (x-i) and (x-(-i)). (The last factor will simplify to (x+i). Our polynomial may also have a constant factor. For now we will call this constant factr "k". So our polynomial is:
All that is left to do is simplify and figure what number k is. To multiply I am going to take advantage of the Associative Property for multiplication:
because, grouped this way, I can use the following patterns
to multiply quickly:
Since this simplifies to:
The last multiplication will have to be done without patterns. Multiplying each term of the trinomial by each term of the binomial we get:
which simplifies as follows:
Now we find k. We will use the given fact of P(2) = 25:
which simplifies as follows:
Dividing by -5 we get:
Now that we have k:
which simplifies to:
which is the desired polynomial.
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