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How do you write a polynomial with zeros 2i, 3i, 1?

1 Answer

A. S. Adikesavan

Apr 24, 2016

$x^{5} - x^{4} + 13 x^{3} - 13 x^{2} + 36 x - 36 = 0$ $x^5-x^4+13x^3-13x^2+36x-36=0$

Explanation:

Complex roots occur in conjugate pairs. So, the roots are $1, \pm 2 i, \pm 3 i,$ $1, +-2i, +-3i,$ . The degree of the polynomial is 5.

Hence, the polynomial equation is

$(x - 1) ((x - 2 i) (x + 2 i)) ((x - 3 i) (x + 3 i)) = 0$ $(x-1)((x-2i)(x+2i))((x-3i)(x+3i))=0$

$(x - 1) (x^{2} + 2^{2}) (x^{2} + 3^{2}) = 0$ $(x-1)(x^2+2^2)(x^2+3^2)=0$

Expanding,

$x^{5} - x^{4} + 13 x^{3} - 13 x^{2} + 36 x - 36 = 0$ $x^5-x^4+13x^3-13x^2+36x-36=0$

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