Question

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.
1) i, 2 - √3

2) 1, 4, 1 + √2

Answer 1

oooh damn ignore all that!!

Answer 2

it is zeros at \(x=i, x=2-\sqrt{3}\) so i am completely wrong, i misread the question

Answer 3

start with
\[(x-i)(x+i)(x-(2-\sqrt{3}))(x-(2+\sqrt{3}))\]

Answer 4

the first product is \[(x-i)(x+i)=x^2+1\] as before
the second product is not as bad as it might look.

Answer 5

you get \[((x-(2+\sqrt{3}))(x-(2-\sqrt{3}))\] do not distribute and multiply, instead think.
first term is clearly \(x^2\)
the constant it \((2+\sqrt{3})(2-\sqrt{3})=2^2-3=4-3=1\)
and the "middle term" is \(-2x-2x=-4x\) as the radicals add to zero.
so your answer is
\[((x-(2+\sqrt{3}))(x-(2-\sqrt{3}))=x^2-4x+1\]

Answer 6

and your final job is to multiply \((x^2+1)(x^2-4x+1)=x^4-4 x^3+2 x^2-4 x+1\)

Answer 7

same idea for the second one
\[(x-1)(x-4)(x-(1+\sqrt{2})(x-(1-\sqrt{2}))\]
do this part \((x-(1+\sqrt{2})(x-(1-\sqrt{2}))\) first