Question

Find the polynomial of least degree, having integral coefficients and leading coefficients equal to 1, with square root of 3 and square root of 2 as a zero.

Answer 1

Each root c has to have an associated factor, x-c.

Since \(\sqrt{3}\) is a root, you must have \((x-\sqrt{3})\) as a factor.

However, in order to have integer coefficients, you must then also have \((x+\sqrt{3})\) as a factor.

so you can start with:
\((x-\sqrt{3})(x+\sqrt{3})\)

Answer 2

Similiarly for \(\sqrt{2}\), you will have to have \((x-\sqrt{2})\) as a factor - and what else?

Multiply it all out, and you'll have your polynomial that meets the specified criteria.

Answer 3

so for the square root of 3.. I'll have an equation of x^2 - 3

Answer 4

Well, you'll have a FACTOR of \((x^2-3)\), yes. (That isn't an equation. :)

Answer 5

then I'll have x^2 - 2

Answer 6

right... now multiply those two factors together....

Answer 7

sorry.. and what next?

Answer 8

\((x^2-3)(x^2-2)\)

Answer 9

Just use FOIL.

Answer 10

so I'll have x^4-5x^2+6

Answer 11

Is that correct?

Answer 12

yup! :)

Answer 13

WOOOH! Thanks a lot! :D

Answer 14

you're welcome. :)