Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions.

Question

n=3

2 and 4i are zeroes

f(-1)=-51

Answer 1

Hello Elly,

First, recall the Conjugate Zeros Theorem:

Let f(x) be a polynomial function with real coefficients. If the complex number a + bi is a zero of f(x), then the conjugate of a + bi is also a zero of f(x).

Since we are given that 4i is a zero, we know that -4i is also a zero. The polynomial function of interest has degree 3, so we know that the zeros of f(x) are 4, 4i, and -4i. Now recall the Factor Theorem:

r is a zero of the polynomial function f(x) if and only if x - r is a factor of f(x).

Therefore, we can write f(x) as

f(x) = A(x - 4)(x - 4i)(x + 4i)

We can now find A using the fact that f(-1) = -85

f(-1) = -85

A(-1 - 4)(-1 - 4i)(-1 + 4i) = -85.

Noting that (-1 - 4i)(-1 + 4i) = 17, this becomes

-85A = -85, so

A = 1.

Hence, f(x) = 1(x - 4)(x -4i)(x + 4i). Multiply out the factors on the right side to get the expanded form.

f(x) = x³ - 4x² + 16x - 64.

Hope that helps! Let me know if you need any further explanation.

William

Answer 2

n=3 means there are three factors that generate zeros for the polynomial.

(x-a)(x-b)(x-c) = 0

you are told two of the three values for a,b,c

(x-4)(x-4i)(x-?) = 0

nice thing about imaginary/complex zeroes is that the come in pairs. if x=2i is a zero, then x=-2i is one as well (look up complex conjugate pairs)

so you now have (x-4)(x-4i)(x+4i)=0

if you plug in x=-1 into the above you would like to get a result of all of that equaling -85.

one catch there could be some constant multiplier in the mix we haven't accounted for since anything time zero is still zero:

k*0=0 (something times zero is zero.)

if (x-4)(x-4i)(x+4i)=0 then k*(x-4)(x-4i)(x+4i)=0 (multiplying zero by k is still going to be zero)

End Game: you need to work with k*(x-4)(x-4i)(x+4i)=0

So let x=-1 and play around to find a k so that the whole thing equals -85

hint: multiply/foil (x-4i)(x+4i) before starting