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
<pre>
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(x) = a(x + 5)(x^2 - 8x + 25)}}} ------ Expanding (x – 4 - 3i)(x - 4 + 3i)
{{{f(2) = a(2 + 5)((2)^2 - 8(2) + 25)}}} -------- 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 ----- {{{(91)/91 = a}}} ----- 1 = a
Therefore, f(x) = a(x + 5)(x – 4 - 3i)(x - 4 + 3i) becomes {{{f(x) = 1(x + 5)(x^2 - 8x + 25)}}}, or {{{highlight_green(highlight_green(f(x) = (x + 5)(x^2 - 8x + 25)))}}}
Expand these polynomials to obtain the 3rd degree polynomial.