Question

N=4; 2i and 3i are zeros f(-1)=50

Answer 1

Solution:- $\text{Let f(x) be any nth degree polynomial with n=4}$

$\text{Given that 2i and 3i are the zeroes of f(x)}$

$\text{so (x-2i) and (x-3i) are factors of f(x)}$

$\text{Since 2i is a zero of f(x) then its conjugate -2i is also a zero of f(x)}$

$\Rightarrow(x+2i) \text{is a factor}\\\text{Similarly, conjugate of 3i is -3i is also a zero of f(x) }\\\Rightarrow(x+3i)\text{is a factor}\\\text{So , }\\f(x)=k(x-2i)(x+2i)(x-3i)(x+3i)\\=k(x^2-(2i)^2)(x^2-(3i)^2)\\=k(x^2-4i^2)(x^2-9i^2)\\=k(x^2+4)(x^2+9)\\=k(x^4+13x^2+36)\\\text{As given}\\f(-1)=50\\\Rightarrow k((-1)^4+13(-1)^2+36)=50\\\Rightarrow k(1+13+36)=50\\\Rightarrow k(50)=50\\\Rightarrow k=1\\\text{So by substituting k=1 in f(x) we get ,}\\f(x)=(x^4+13x^2+36)$