How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are i, -3i, 3i?

Question

1442 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-11T00:15:12

$x^{3} - x^{2} i + 9 x - 9 i = 0$ $x^3-x^2i+9x-9i=0$

When we talk about the zeros of a polynomial, we can express them as $x = the zero$ $x="the zero"$ . So with the terms listed above, we can say:

$x = i, - 3 i, 3 i$ $x=i, -3i, 3i$ and we can therefore say:

$x - i = 0, x + 3 i = 0, x - 3 i = 0$ $x-i=0, x+3i=0, x-3i=0$

which leads to:

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

Expanding and collecting like terms for coefficients in the cubic

$x^{3} - x^{2} i + 9 x - 9 i = 0$ $x^3-x^2i+9x-9i=0$ .

Also, we can use the form

$x^{3}$ $x^3$ -

(sum of the roots)x^2+

sum of the products of the roots, taken two at a time)x-

product of the roots

= 0.

Here, it is

$x^{3} - (i + 3 i - 3 i) x^{2} + (3 - 3 + 9) x - (9 i) = 0$ $x^3-(i+3i-3i)x^2+(3-3+9)x-(9i)=0$ . Simplifying,

$x^{3} - x^{2} i + 9 x - 9 i = 0$ $x^3-x^2i+9x-9i=0$