3 years ago

Please help brainliest for best

Mathematics

1 answer:

V125BC [204]3 years ago

6 0

Hdjsjsjzjjzjznzjzjhxjxj

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

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

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

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

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