Question 880484: Find the nth degree polynomial with real coefficients satisfying the given conditions. n=3; 1 and 5i are zeros,f(-1) = -104
I have no idea how to do this, so can whomever help with it step-by-step. I want to analyze every last step it takes to understand this question. What do I have to do?
Found 3 solutions by ewatrrr, josgarithmetic, MathTherapy:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! 1 and 5i are zeros. The 3 zeroes are 1, 5i, -5i
f(x)= a(x-1)(x-5i)(x+5i)
f(x)= a(x-1)(x^2 + 25)
f(-1) = a(-2)(1+25) = -52a = -104, a = 2
f(x)= 2(x-1)(x^2 + 25) expand if You wish
Answer by josgarithmetic(39620)  (Show Source):
You can put this solution on YOUR website!  , and you are given the zeros, meaning  . You have only one more zero to use and you should understand that complex zeros with imaginary parts come in "conjugate pairs". This means, since you are given 5i as one of the zeros, -5i is ALSO a zero.
Handling just the needed, and now known zeros, the binomial factors for the function are these combined:  .
Simplify that expression.
f will also have another constant factor, k, needed in order to match the condition, f(-1)=-104. You could manage that part like  , in which the product shown in previous paragraph is that polynomial.
You may be able to manage the details and the rest of finding f(x).
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Find the nth degree polynomial with real coefficients satisfying the given conditions. n=3; 1 and 5i are zeros,f(-1) = -104
I have no idea how to do this, so can whomever help with it step-by-step. I want to analyze every last step it takes to understand this question. What do I have to do?
n = 3 indicates that there are 3 zeroes, or 3 solutions to the equation. Two of the 3 zeroes or solutions
are 1 and 5i, but complex numbers such as 5i come in CONJUGATE PAIRS. The conjugate of 5i is its opposite, – 5i
Now, we have 3 zeroes/solutions, as follows: x = 1; x = 5i, and x = - 5i
x = 1____x – 1 = 0
x = 5i____x – 5i = 0
x = - 5i____x + 5i = 0
We now have the following:
f(x) = a(x – 1)(x – 5i)(x + 5i)
 ------ FOILing/Expanding (x – 5i)(x + 5i)
 -------- Substituting – 1 for x to determine value of a
f(- 1) = a(- 2)(1 + 25)
f(- 1) = - 2a(26)
f(- 1) = - 52a
Now, since f(- 1) = - 104, then we can say that: - 104 = - 52a -----  ----- 2 = a
Therefore, f(x) = a(x – 1)(x – 5i)(x + 5i) becomes  , or 
FOIL/Expand these binomials to obtain the 3rd degree polynomial.
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