Is it possible to factor #y= x^5+2x^4+x^3+4x^2+5x+2 #? If so, what are the factors?

Let #f(x) = x^5+2x^4+x^3+4x^2+5x+2#

By the rational root theorem, the only possible rational roots of #f(x) = 0# are #+-1#, #+-2#.

Trying these we find:

#f(-2) = -32+32-8+16-10+2 = 0#

So #(x+2)# is a factor:

#x^5+2x^4+x^3+4x^2+5x+2 = (x+2)(x^4+x^2+2x+1)#

Next, since the remaining quartic factor has no #x^3# term, it will factor into two quadratics with opposing middle coefficients:

#x^4+x^2+2x+1#

#= (x^2+ax+b)(x^2-ax+c)#

#=x^4+(b+c-a^2)x^2+a(c-b)x+bc#

Equating coefficients and rearranging slightly, we get:

#b+c = a^2+1#

#c-b = 2/a#

#bc = 1#

So:

#(a^2 + 1)^2 = (b+c)^2 = (c-b)^2 + 4bc = (2/a)^2 + 4#

That is:

#(a^2)^2+2(a^2)+1 = 4/a^2 + 4#

Multiplying through by #a^2# and rearranging a bit, that becomes:

#(a^2)^3+2(a^2)^2-3(a^2)-4 = 0#

One of the solutions to this cubic in #a^2# is #a^2 = -1#, but we would like to find a root that gives us a Real value for #a#, so let's factor this cubic and try again:

#(a^2)^3+2(a^2)^2-3(a^2)-4 = (a^2+1)((a^2)^2+a^2-4)#

Hence two further roots:

#a^2 = (-1+-sqrt(17))/2#

In particular, #a^2 = (sqrt(17)-1)/2#, giving us:

#a = +-sqrt((sqrt(17)-1)/2)#

We can use the positive square root, since the derivation is symmetric.

So, adding #b+c = a^2+1# and #c - b = 2/a#

we find:

#2c = a^2+1+2/a#

#=(sqrt(17)-1)/2 + 1 + 2/sqrt((sqrt(17)-1)/2)#

So:

#c = (sqrt(17)-1)/4 + 1/2 + 1/sqrt((sqrt(17)-1)/2)#

#= (sqrt(17)+1)/4 + sqrt((sqrt(17)-1)/2)/((sqrt(17)-1)/2)#

#= (sqrt(17)+1)/4 + (2sqrt((sqrt(17)-1)/2))/(sqrt(17)-1)#

#= (sqrt(17)+1)/4 + (2sqrt((sqrt(17)-1)/2)(sqrt(17)+1))/16#

#= ((2+sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8#

Similarly:

#b = ((2-sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8#

Putting it together:

#x^4+x^2+2x+1#

#=(x^2+sqrt((sqrt(17)-1)/2)x+((2-sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8)#

#*(x^2-sqrt((sqrt(17)-1)/2)x+((2+sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8)#

Ouch!