SOLUTION: Find an nth degree polynomial function with real coefficients satisfying the given conditions. N=3; -5 and 4 + 3i are zeros; f(2) = 91 My math book doesnt give an example for this

Algebra ->  Rational-functions -> SOLUTION: Find an nth degree polynomial function with real coefficients satisfying the given conditions. N=3; -5 and 4 + 3i are zeros; f(2) = 91 My math book doesnt give an example for this      Log On


   

Question 880837: Find an nth degree polynomial function with real coefficients satisfying the given conditions. N=3; -5 and 4 + 3i are zeros; f(2) = 91
My math book doesnt give an example for this problem so im confused, i just know a=-1
And the problem should end up as x^3-3x^2-15x+125
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
I just did that one yesterday. You already sent me a Thanks note about it. I just did not perform the last multiplication step to put the function into general form but instead finished my solution still in factored form.

This should be the same:
http://www.algebra.com/algebra/homework/Rational-functions/Rational-functions.faq.question.880683.html

You should be sure you can follow my solution. Notice, my leading coefficient was different. I doubt that I made any mistake in any steps; but it is possible.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Find an nth degree polynomial function with real coefficients satisfying the given conditions. N=3; -5 and 4 + 3i are zeros; f(2) = 91
My math book doesnt give an example for this problem so im confused, i just know a=-1
And the problem should end up as x^3-3x^2-15x+125


n = 3 indicates that there are 3 zeroes, or 3 solutions to the equation. Two of the 3 zeroes or solutions

are - 5 and 4 + 3i, but complex numbers such as 4 + 3i come in CONJUGATE PAIRS. The conjugate of 4 + 3i is: 4 – 3i



Now, we have 3 zeroes/solutions, as follows: x = - 5; x = 4 + 3i, and x = 4 - 3i

x =  - 5________x + 5      = 0

x =    4 + 3i___x – 4 - 3i = 0

x =    4 - 3i___x - 4 + 3i = 0



We now have the following:

f(x) = a(x + 5)(x – 4 - 3i)(x - 4 + 3i)

f%28x%29+=+a%28x+%2B+5%29%28x%5E2+-+8x+%2B+25%29 ------ Expanding (x – 4 - 3i)(x - 4 + 3i)

f%282%29+=+a%282+%2B+5%29%28%282%29%5E2+-+8%282%29+%2B+25%29 -------- Substituting 2 for x to determine value of a

f(2) = a(7)(4 - 16 + 25)

f(2) = 7a(13)

f(2) = 91a



Now, since f(2) = 91, then we can say that: 91 = 91a ----- %2891%29%2F91+=+a ----- 1 = a

Therefore, f(x) = a(x + 5)(x – 4 - 3i)(x - 4 + 3i) becomes f%28x%29+=+1%28x+%2B+5%29%28x%5E2+-+8x+%2B+25%29, or highlight_green%28highlight_green%28f%28x%29+=+%28x+%2B+5%29%28x%5E2+-+8x+%2B+25%29%29%29

Expand these polynomials to obtain the 3rd degree polynomial.