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 Math Help > Basic Math > Mental Math > Clever Substitutions

### "Factoring"

The word factoring is in quotes, because these substitutions are a bit more off-the-wall than just factoring.  You'll see:

1+x4 = (x2-sqrt(2)x+1)(x2+sqrt(2)x+1)

### Fourth degree polynomial hints

Reciprocal equation

x4 + ax3 + bx2 + ax + 1 = 0
x2 + ax + b + a/x + 1/x2 = 0
x2 + 1/x2 + a(x+1/x) + b = 0
(x+1/x)2 + a(x+1/x) + b - 2 = 0

More generally, if if B=A2/4+2C/A then x4 + Ax3 + Bx2 + Cx + D = 0 can be expressed as

(x2 + ax + b)2 + c(x2 + ax + b) + d = 0, which expands as
x4 + 2ax3 + (a2+2b+c)x2 + (2b+c)ax + b2+bc+d = 0

The four solutions, then, are found using the quadratic formula twice:

x2 + ax + b = (-c ï¿½ sqrt(c2-4d))/2
x = -a ±sqrt(a2-4b-2c ±2 sqrt(c2-4d))/2

A = 2a
B = a2+2b+c
C = a(2b+c)
D = b2+bc+d

a=A/2
b=(Cï¿½sqrt(dA2-DA2+C2))/A
c=(-2) (ï¿½sqrt(dA2-DA2+C2)/A
d=arbitrary

dA2-DA2+C2=0, so d=D-C2/A2, 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):

x4+2x3+7x2+6x+8=0

Notice that B=A2/4+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=(C-sqrt(dA2-DA2+C2))/A=0
c=(-2) (-sqrt(dA2-DA2+C2)/A=6

So (x2+x)2+6(x2+x)+8=0

But, if you like, you can let d=0, which gives you a=1, b=2, c=2, and thus

(x2+x+2)2+2(x2+x+2)=0

Or, as I suggest above, d=D-C2/A2 = -1 gives you

(x2+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
(x2+ax+b)2 = x4+2ax3+(a2+2b)x2+2abx+b2

### The Quaternion Identity

So called because if u=a-bi-cj-dk and U=A+Bi+Cj+Dk, |uU| = |u||U|.  Squaring both sides gives:

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 Complex Product Identity

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:

(a2+b2)(A2+B2) = (aA+bB)2+(aB-bA)2

A variant of the Complex Product Identity shows that the product of two numbers, each of which has the form a2+3b2 can also be expressed in the form a2+3b2.

(a2+3b2)(A2+3B2)=(aA-3bB)2+3(aB+bA)2