How do you write a polynomial with zeros i, -3i, 3i and leading coefficient 1?

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Answer 1 · 2016-10-08T08:14:43

The polynomial with zeros $i$ $i$ , $- 3 i$ $-3i$ and $3 i$ $3i$

is $x^{3} - i x^{2} + 9 x - 9 i$ $x^3-ix^2+9x-9i$

A polynomial with zeros $a$ $a$ , $b$ $b$ and $c$ $c$ is

$(x - a) (x - b) (x - c)$ $(x-a)(x-b)(x-c)$

Hence a polynomial with zeros $i$ $i$ , $- 3 i$ $-3i$ and $3 i$ $3i$ is

$(x - i) (x - (- 3 i)) (x - 3 i)$ $(x-i)(x-(-3i))(x-3i)$

= $(x - i) (x + 3 i) (x - 3 i)$ $(x-i)(x+3i)(x-3i)$

= $(x - i) (x^{2} + 9)$ $(x-i)(x^2+9)$

= $x^{3} - i x^{2} + 9 x - 9 i$ $x^3-ix^2+9x-9i$

Here leading coefficient (of $x^{3}$ $x^3$ ) is already one.

Hence, the polynomial with zeros $i$ $i$ , $- 3 i$ $-3i$ and $3 i$ $3i$ is $x^{3} - i x^{2} + 9 x - 9 i$ $x^3-ix^2+9x-9i$