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Factor quartics into a pair of symmetric quadratics

Opposite linear terms create a difference of squares

A symmetric pair $(x^{2} + a x + b) (x^{2} - a x + b)$ equals $(x^{2} + b)^{2} - a^{2} x^{2}$ . Expand this shorter expression and compare the constant and quadratic coefficients.

Problem

Example 1. If $x^{4} - 3 x^{2} + 1 = (x^{2} + a x - b) (x^{2} - a x - b)$ with $a, b > 0$ , find $ab$ . Example 2. If $x^{4} + x^{2} + 25 = (x^{2} + a x + b) (x^{2} - a x + b)$ with $a, b > 0$ , find $a^{2} + b^{2}$ .

Answer

Example 1:

a = b = 1

, so

ab = 1

. Example 2:

a = 3

b = 5

, so

a^{2} + b^{2} = 34

The sign of b inside the quadratics differs between the two examples.

Step-by-step solution

1. Expand the first symmetric pair

Using a difference of squares gives

(x^{2} - b)^{2} - a^{2} x^{2} = x^{4} - (2 b + a^{2}) x^{2} + b^{2} .

Compare this with $x^{4} - 3 x^{2} + 1$ . Since $b^{2} = 1$ and $b > 0$ , $b = 1$ . Then $2 b + a^{2} = 3$ gives $a = 1$ , so $ab = 1$ .

2. Expand the second pair

Here the product is

(x^{2} + b)^{2} - a^{2} x^{2} = x^{4} + (2 b - a^{2}) x^{2} + b^{2} .

Comparing with $x^{4} + x^{2} + 25$ gives $b^{2} = 25$ , hence $b = 5$ . Then $2 b - a^{2} = 1$ gives $a^{2} = 9$ and $a = 3$ . Therefore $a^{2} + b^{2} = 9 + 25 = 34$ .

3. Check the factorizations

The products are $(x^{2} + x - 1) (x^{2} - x - 1)$ and $(x^{2} + 3 x + 5) (x^{2} - 3 x + 5)$ . Expanding them restores the two original quartics.

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