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

Question

n=​3;

−2

and

7+5i

are​ zeros;

Answer 1

-2, 7+5i, 7-5i are the 3 zeros, the imaginary zeros always come in conjugate pairs so if 7+5i is a zero then so too is 7-5i a zero

x=-2, x+2=0

x=7+5i and 7-5i

(x-(7+5i)(x-(7-5i))= (x-7-5i)(x-7+5i) = x^2 -14x+49+25

x^2 -14x+74

corresponding factors are

(x+2)(x^2-14x+74)

= x^3 -12x^2 +46x + 128

or any multiple such as double

each term = 2x^3 -24x^2 + 92x + 256

also have the same required zeros and degree 3

Answer 2

Answer 3

So you want y = (x- [whatever one of the given zeros are] ) * ( x- [whatever one of the given zeros are] ) * (x- [whatever one of the given zeros are] ).

Since the zeros are -2 and 7+5i, you have:

y = ( x - [-2] ) * (x - [5+7i] ) * (x - [one of the given zeros])

so y = (x+2)*(x-5-7i)*((x - [one of the given zeros]) so just pick one of the zeros you already used.

so either y = (x+2)*(x-5-7i)*(x+2) OR (x+2)*(x-5-7i)*(x-5-7i)

Then make sure to simply to: either (x+2)^2 * (x-5-7i) OR y= (x+2)*(x-5-7i)^2