How do you write the polynomial function given the following zeros: -5i, -i, i, 2i?

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Answer 1 · 2015-09-17T18:43:12

It is $P (x) = (x + 5 i) (x + i) (x - i) (x - 2 i) = x^{4} + 3 i x^{3} + 11 x^{2} + 3 i x + 10$ $P(x)=(x+5i)(x+i)(x-i)(x-2i)=x^4+3ix^3+11x^2+3ix+10$

where $i$ $i$ is the imaginery unit ( $i^{2} = - 1$ $i^2=-1$ )

Answer 2 · 2015-09-17T19:52:28

If the polynomial also has Real coefficients, then it also has zeroes $5 i$ $5i$ and $- 2 i$ $-2i$

The simplest such polynomial is:

$f (x) = (x - 5 i) (x + 5 i) (x - i) (x + i) (x - 2 i) (x + 2 i)$ $f(x) = (x-5i)(x+5i)(x-i)(x+i)(x-2i)(x+2i)$

$= x^{6} + 30 x^{4} + 129 x^{2} + 100$ $=x^6+30x^4+129x^2+100$

The Complex zeroes of polynomials with Real coefficients always occur in conjugate pairs.

So in our example, since $- 5 i$ $-5i$ and $2 i$ $2i$ are zeroes, so are $5 i$ $5i$ and $- 2 i$ $-2i$ .

Hence:

$f (x) = (x - 5 i) (x + 5 i) (x - i) (x + i) (x - 2 i) (x + 2 i)$ $f(x) = (x-5i)(x+5i)(x-i)(x+i)(x-2i)(x+2i)$

$= (x^{2} + 25) (x^{2} + 1) (x^{2} + 4)$ $= (x^2+25)(x^2+1)(x^2+4)$

$= x^{6} + (25 + 1 + 4) x^{4} + (1 \cdot 4 + 25 \cdot 4 + 25 \cdot 1) x^{2} + (25 \cdot 1 \cdot 4)$ $=x^6+(25+1+4)x^4+(1*4+25*4+25*1)x^2+(25*1*4)$

$= x^{6} + 30 x^{4} + 129 x^{2} + 100$ $=x^6+30x^4+129x^2+100$

Any polynomial in $x$ $x$ with these roots will be a multiple (scalar or polynomial) of this $f (x)$ $f(x)$ .