How do you factor # x^4 + 4x^3  22x^2 + 4x + 16#?
1 Answer
Answer:
#x^4+4x^322x^2+4x+16#
#=(x+3sqrt(17))(x+3+sqrt(17))(x1sqrt(3))(x1+sqrt(3))#
Explanation:
We can try the rational roots theorem, which tells us that any rational zeros of
That means that the only possible rational zeros are:
#+1, +2, +4, +8, +16#
None of these works, so
How about quadratic factors?
Is there a pair of quadratic factors with integer coefficients?
We can simplify the problem by making a linear substitution.
#x^4+4x^322x^2+4x+16#
#=(x+1)^428(x+1)^2+56(x+1)13#
#=t^428t^2+56t13#
where
If this simplified quartic has quadratic factors with integer coefficients then they must take one of the following forms:

#(t^2+at+13)(t^2at1)# 
#(t^2+at13)(t^2at+1)#
With a little comparing of coefficients we find:
#t^428t^2+56t13 = (t^2+4t13)(t^24t+1)#
Hence:
#x^4+4x^322x^2+4x+16#
#=((x+1)^2+4(x+1)13)((x+1)^24(x+1)+1)#
#=(x^2+6x8)(x^22x2)#
#=((x+3)^217)((x1)^23)#
#=(x+3sqrt(17))(x+3+sqrt(17))(x1sqrt(3))(x1+sqrt(3))#