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

Question

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.

n=3

2 and 5i are zeros

f(1)=-52

Answer 1

Imaginary roots always occur in pairs so the equation is y=a(x-2)(x+5i)(x-5i)

Since f(1)= -52. -52 = a(1-2)(1+5i)(1-5i)

-52= -a(1+25)

-52= -a(26)

a=2

So equation is y=(x-2)(x+5i)(x-5i) This is factored form. Can multiply this out to put in standard form

y=2(x-2)(x^2+25)

y=2(x^3-2x^2+25x-50)

Answer 2

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.

1. n=3 ; 2 and 5i are zeros; f(1)=-52

Since f(x) has real coefficients 5i is a root, so is -5i

So, 2, 5i, and -5i are roots

Since f(x) has degree 3, there can't be more than 3 distinct roots.

f(x) = a(x-2)(x-5i)(x+5i)

f(x) = a(x-2)(x² +25)

f(1) = -52, so

-52 = a(1-2)(1²+25)

-52 = a(-1)(26)

-52/-26 = a

2 = a

f(x) = 2(x-2)(x² +25)

You can use your graphing calculator (or desmos) to verify the real zero at x = 2

I hope this helps you!