Question

what are the sets of coincident zeros are possessed by polynomial x^4-2x^2+1​

Answer 1

Q]what are the sets of coincident zeros are possessed by polynomial x^4-2x^2+1

ANS ]

⭕⚫[x ^4 - 2x ^2 +1]

⚫

$(x { }^{2}) {}^{2} - 1x {}^{2} - 1x {}^{2} + 1$

$x {}^{2} (x {}^{2} - 1) - 1(x {}^{2} - 1)$

$(x {}^{2} - 1)(x {}^{2} - 1)$

$x = \sqrt{1} \: or x = \sqrt{1}$

⚫THEREFORE

$x = 1$

Answer 2

Answer:

+1, -1

Step-by-step explanation:

let x^2 be y

therefore, the polynomial becomes p(x) =  y^2 - 2y + 1

=> p(x) = (y-1)^2

on replacing y by x^2

we get p(x) = (x^2 - 1)(x^2 - 1)

so x = +1, -1, +1, -1

so the coincident zeroes are +1, -1