Question

find an nth-degree polynomial function with real coefficients satisfying the give conditions
n=3;-4 and 6+5i are zero; f(1)=250

Answer 1

So it's a degree 3.

Answer 2

If \(6+5i\) is a root than so is \(6-5i\)

Answer 3

So you know it is of the form\[
c(x-(-4))(x-(6+5i))(x-(6-5i)) = c(x+4)(x^2-12x+61)
\]

Answer 4

You can solve for \(c\) by using: \[
250=c((1)+4)((1)^2-12(1)+61)
\]

Answer 5

hey how can u solve (x-(6+5i))(x-(6-5i))=(x^2-12x+61) can u explain more thank

Answer 6

Solve by multiplication and comparison. The coefficients on the two resulting polynomials must be equal.