Question

Find the polynomial function with roots 11 and 2i.

Answer 1

for a polynomial of say
\(
x^2+2x+1
\)
what would you say are its roots?

Answer 2

do you even know what a 'root' of an expression is? :|

Answer 3

2 and 1 I think

Answer 4

well, 1 :|, because \(
x^2+2x+1
\) = (x+1)(x+1)=0, x=1 and x=1, so it has 2 roots, both are 1 in this case

Answer 5

a root of a polynomial means, if you equate it to 0, that is \(
x^2+2x+1
=y \implies x^2+2x+1=0\), that is, the function is touching the X-axis, what values does X have

Answer 6

you usually end up with one or more binomials, like the one above "x+1)(x+1)=0", so in this case your function where 'y=0', you'd have 2 binomials, like
$$
\boldsymbol{i=\sqrt{-1}}\\
\color{blue}{(x-11)(x-2\sqrt{-1})=0}\\
x=11, x=2\sqrt{-1} = 2i
$$

Answer 7

if you multiply/expand those 2 binomials, you'd get your equation :)