p(x)=3x4+6x3−2x2−10x−5
Since the two zeroes are √53 and −√53 ,
∴(x−√53)(x+√53)=(x2−53) is a factor of 3x4+6x3−2x2−10x−5.
Therefore, we divide the given polynomial by x2−53.
We factorize 3x2+6x+3=0
=3(x2+2x+1)=0
=(x+1)2=0
Therefore, its zero is given by x + 1 = 0.
x = -1
As it has the term (x+1)2 , therefore, there will be 2 zeroes at x = - 1.
Hence, the zeroes of the given polynomial are √53 and −√53 and – 1.