Nth-degree polynomial

Question

N=4

i and 4i are zeros

f(-1)=68

Answer 1

Then -i and -4i are zeros if the coefficients are real

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

= k ( x^2+1)(x^2 + 16)

= k( x^4 + 17x^2 + 16)

68 = k(1 + 17 + 16)

68 = k(34)

k=2

f(x) = 2(x^4 + 17x^2 + 16)

Answer 2

Since i is a root, so is -i. Since 4i is a root, so is -4i.

f(x) = A(x - i)[x - (-i)](x - 4i)[x - (-4i)]

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

= A(x² + 1)(x² + 16)

= A(x⁴ + 17x² + 16)

Since f(-1) = 68, 34A = 68. So, A = 2.

f(x) = 2(x⁴ + 17x² + 16)

f(x) = 2x⁴ + 34x² + 32