#x^4-6x^2+1=0#
This is clearly a quadratic equation in #x^2#, so
#x^2=(-(-6)+-sqrt((-6)^2-4*1*1))/(2*1)=3+-2sqrt(2)#
So the four solutions to this quartic equation are
#x=+-sqrt(3+-2sqrt(2))=+-(1+-sqrt(2))#
Hence the four factors of the given polynomial are
#(x-1-sqrt2)#, #(x-1+sqrt2)#, #(x+1-sqrt2)#, #(x+1+sqrt2)#
So,
#(x-1-sqrt2) * (x-1+sqrt2) * (x+1-sqrt2) * (x+1+sqrt2)=x^4-6x^2+1#
Hence the linear factors are not easily deducible.
Alternate method to factor the given polynomial into two quadratics#:-#
Since the leading coefficient in the equation is one, we can set this equation to be a product of two quadratics such that
#x^4-6x^2+1=(x^2+ax+b) * (x^2+cx+d)#
#=>x^4-6x^2+1=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd#
Comparing the L.H.S. and R.H.S. of this equation, we get
#a+c=0#
#ac+b+d=-6#
#ad+bc=0#
#bd=1#
Solving these equations simultaneously we get
#a=-2#, #b=-1#, #c=2#, #d=-1#
#(#There are other sets of the values for #a,b,c,d#. But I am not writing them here.#)#
#=>x^4-6x^2+1=(x^2-2x-1)*(x^2+2x-1)#
#[#Other possible factors are
#x^4-6x^2+1=(x^2-2sqrt(2)x+1)*(x^2+2sqrt(2)x+1)#, and
#x^4-6x^2+1=(x^2-3-2sqrt2)*(x^2-3+2sqrt2)# #]#
