Question

Find a cubic polynomial with zeros 2 RADICAL 2 and NEGATIVE RADICAL 2.

For future reference how would I quickly do this? Trial and error will not cut it!

Answer 1

Write in equation tab the roots

Answer 2

Could you please explain, what is RADICAL 2

Answer 3

\[-\sqrt{2}\]\[\sqrt{2}\]\[2\]

Answer 4

\[\sqrt{2}\] is RADICAL 2?

Answer 5

Yes.

Answer 6

ok, then the polynomial can be written as \[(x-2)(x-\sqrt{2})(x+{\sqrt{2})}\]
= x3 -2x2 - 2x + 4

Answer 7

does it answer your question?

Answer 8

I'm trying to find a polynomial with at least a degree of three it also has to have three the zeros I mentioned in the previous reply.

Answer 9

yeah, this polynomial has so if you equate the polynomial with 0 and try to find the roots they will come as 2, RADICAL 2 and NEGATIVE RADICAL 2.

\[x ^{3}-2x ^{2}-2x+4 = 0\]
Then (\[(x-2)(x ^{2}-2) = 0\]
Hence, x=2 or x^2 = 2 => x= 2, or x - RADICAlL 2 or x= NEGATIVE RADICAL 2

Answer 10

It has a degree of three or else three roots would not have existed

Answer 11

and also highest power of x is 3

Answer 12

Alrighty. Now how did you find the answer so quickly? Is there a formula or method of doing this?

Answer 13

yeah, what I did is the solution of equation by factorization in the backward way, I have subtracted (x-2) and (x-sqrt2)[positive radical](x+sqrt2)[negetive radical] and multiplied them.

You do exactly opposite when you factorise the polynomial and try to equate each factor with 0 for solution

Answer 14

Haha I see. so if I wanted to do a polynomial that had zeros of -2 and 1 it would be.
x+2*x-1
Right?

Answer 15

yes correct (x+2)(x-1) = x^2 + x -2