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; 4 and 2 i are zeros; f(1) = 15 f(x)=0

Answer 1

for a n degree polynomial there will n zeros . this include both real and imaginary zeros .

here degree of polynomial is 3 . so it will have 3 zeros .

if c is a zero then x-c will be a factor of this polynomial .

if all coefishent of a polynomial are real and it has a imaginary zero , then its conjugate also will be a zero .

here 2i is a imaginary zero . given all coefishent are real . so another zero is -2i ( conjugate of 2i) .

so all zeros of this function are 4, 2i , -2i

so polynomial has the factors x-4 , x-2i and x+2i

so

f(x)= a (x-4)(x-2i)(x+2i)

now expand . we know (a+bi)(a-bi)= a2+b2 . so (x-2i)(x+2i)= x2+4

f(x)= a(x-4)(x2+4)

f(1)= 15 . apply x=1 to above equation

f(1)= a(1-4) (1+4)=15

a*-3*5=15

-15a=15

so a=-1

so

f(x)= -(x-4)(x2+2) = -x3+4x2-4x+16

so answer is

f (x) = -x} + 4x² - 4x + 16