
The word factoring is in quotes, because these substitutions are a bit more offthewall than just factoring. You'll see:
1+x^{4} = (x^{2}sqrt(2)x+1)(x^{2}+sqrt(2)x+1)
Reciprocal equation
x^{4} + ax^{3} + bx^{2} + ax + 1 = 0
x^{2} + ax + b + a/x + 1/x^{2} = 0
x^{2} + 1/x^{2} + a(x+1/x) + b = 0
(x+1/x)^{2} + a(x+1/x) + b  2 = 0Embedded quadratics
More generally, if if B=A^{2}/4+2C/A then x^{4} + Ax^{3} + Bx^{2} + Cx + D = 0 can be expressed as
(x^{2} + ax + b)^{2} + c(x^{2} + ax + b) + d = 0, which expands as
x^{4} + 2ax^{3} + (a^{2}+2b+c)x^{2} + (2b+c)ax + b^{2}+bc+d = 0The four solutions, then, are found using the quadratic formula twice:
x^{2} + ax + b = (c ï¿½ sqrt(c^{2}4d))/2
x = a ±sqrt(a^{2}4b2c ±2 sqrt(c^{2}4d))/2A = 2a
B = a^{2}+2b+c
C = a(2b+c)
D = b^{2}+bc+da=A/2
b=(Cï¿½sqrt(dA^{2}DA^{2}+C^{2}))/A
c=(2) (ï¿½sqrt(dA^{2}DA^{2}+C^{2})/A
d=arbitrarySince d is arbitrary, how about this for d?
dA^{2}DA^{2}+C^{2}=0, so d=DC^{2}/A^{2}, and then c=0 (so the method just completes the square, see below) and b=C/A.Or, you can let d=0, or you can let d=D. Or any other value you choose!
Example (from the fourth puzzle reference, below):
x^{4}+2x^{3}+7x^{2}+6x+8=0
Notice that B=A^{2}/4+2C/A, so this problem meets the criterion for solving it this way. The solution offered lets d=D=8, and so
a=A/2=1
b=(Csqrt(dA^{2}DA^{2}+C^{2}))/A=0
c=(2) (sqrt(dA^{2}DA^{2}+C^{2})/A=6So (x^{2}+x)^{2}+6(x^{2}+x)+8=0
But, if you like, you can let d=0, which gives you a=1, b=2, c=2, and thus
(x^{2}+x+2)^{2}+2(x^{2}+x+2)=0
Or, as I suggest above, d=DC^{2}/A^{2} = 1 gives you
(x^{2}+x+3)^{2}1=0
and, in fact, this is a whole lot less clever (read: more straightforward) than the solution that is offered, because it's done just by completing the square, since
(x^{2}+ax+b)^{2} = x^{4}+2ax^{3}+(a^{2}+2b)x^{2}+2abx+b^{2}
So called because if u=abicjdk and U=A+Bi+Cj+Dk, uU = uU. Squaring both sides gives:
(a^{2}+b^{2}+c^{2}+d^{2})(A^{2}+B^{2}+C^{2}+D^{2}) = (aA+bB+cC+dD)^{2}+(aBbA+cDdC)^{2}+(aCcA+dBbD)^{2}+(aDdA+bCcB)^{2}
Thus, the product of any two numbers, each of which can be expressed as the sum of four squares, can also be expressed as the sum of four squares.
The product of any two numbers, each of which can be expressed as the sum of two squares, can also be expressed as the sum of two squares:
(a^{2}+b^{2})(A^{2}+B^{2}) = (aA+bB)^{2}+(aBbA)^{2}
A variant of the Complex Product Identity shows that the product of two numbers, each of which has the form a^{2}+3b^{2} can also be expressed in the form a^{2}+3b^{2}.
(a^{2}+3b^{2})(A^{2}+3B^{2})=(aA3bB)^{2}+3(aB+bA)^{2}
gifted.hkedcity.net/Gifted/Download/notes/phase3/basic/Forth%20Lecture_20070210.pdf (mirror), which contains a number of useful hints for solving fourthdegree polynomials.
Integral of sqrt(tan(x))  1+x^{4} = (x^{2}sqrt(2)x+1)(x^{2}+sqrt(2)x+1)
Sums of Squares  proof that any number can be expressed as the sum of four squares, using the Quaternion and Complex Product identities.
Trig Equivalences and Table of Integrals give examples of many more clever substitutions.
Factoring  a tutorial for basic factoring
The webmaster and author of this Math Help site is Graeme McRae.