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How do you write a polynomial in standard form given zeros 1 - 2i and 1 + 2i?

1 Answer

Apr 2, 2016

$f (x) = x^{2} - 2 x + 5$ $f(x)=x^2-2x+5$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

Use this with $a = (x - 1)$ $a=(x-1)$ and $b = 2 i$ $b=2i$ as follows:

$f (x) = (x - 1 - 2 i) (x - 1 + 2 i)$ $f(x) = (x-1-2i)(x-1+2i)$

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

$= {(x - 1)}^{2} - {(2 i)}^{2}$ $=(x-1)^2-(2i)^2$

$= x^{2} - 2 x + 1 + 4$ $=x^2-2x+1+4$

$= x^{2} - 2 x + 5$ $=x^2-2x+5$

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

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