Question

find zeroes of polynomial of 2x^3-4x-x^2+2 it's two zeroes are -√2 and √2​

Answer 1

Answer:

Step-by-step explanation:

Since two zeroes are √2 and √−2 , ( x −√2 )(x+√2 ) = x²–2 is a  factor of the given polynomial. Now, we divide the given polynomial by x²–2.

the answer will come out to be 2x²–3x + 1

So, 2x^4 –3x³–3x² +6x –2 = (x²– 2)(2x²–3x + 1).

So, its zeroes  are given by x = 1 /2 and x = 1. Therefore, the zeroes of the given polynomial are 1/2, 1,√2, and √-2

Answer 2

$\sf \pmb{Answer :}$

Step-by-step explanation:

Since two zeroes are √2 and √−2 , ( x −√2 )(x+√2 ) = x²–2 is a  factor of the given polynomial. Now, we divide the given polynomial by x²–2.

the answer will come out to be 2x²–3x + 1

So, 2x^4

–3x³–3x²

+6x –2 = (x²– 2)(2x²–3x + 1).

So, its zeroes  are given by x =

1

/2 and x = 1. Therefore, the zeroes of the given polynomial are

1/2, 1,√2, and √-2