Question

How do you factor the expression `x^4-13x^2+36`?

Answer 1

`x^4-13x^2+36=(x-3)(x+3)(x-2)(x+2)`

Explanation:

We will use the difference of squares identity which can be written:

`a^2-b^2=(a-b)(a+b)`

First, notice that this quartic is a quadratic in `x^2`. Note also that `36 = 9xx4` and `13 = 9+4`, so we find:

`x^4-13x^2+36`

`=(x^2)^2-13(x^2)+36`

`=(x^2-9)(x^2-4)`

`=(x^2-3^2)(x^2-2^2)`

`=(x-3)(x+3)(x-2)(x+2)`

Answer 2

Use an alternative method to find:

`x^4-13x^2+36 = (x-3)(x-2)(x+3)(x+2)`

Explanation:

Another method of factoring `x^4-13x^2+36` goes as follows:

Note that `x^4 = (x^2)^2` and `36 = 6^2` are both perfect squares.

Note also that there are no terms in `x^3` or `x`.

Consider:

`(x^2-kx+6)(x^2+kx+6)=x^4+(12-k^2)x^2+36`

The opposite middle terms `-kx` and `+kx` cause the `x^3` and `x` terms to cancel out.

If `(12-k^2)x^2 = -13x^2` then we find `k^2=25` so `k=+-5` and our original quartic factors as:

`x^4-13x^2+36 = (x^2-5x+6)(x^2+5x+6)`

In our example, each of the remaining quadratic factors is reducible to linear factors with Real coefficients:

`(x^2-5x+6)(x^2+5x+6) = (x-3)(x-2)(x+3)(x+2)`

Note however, that in general we may often be able to factor a quartic with no `x^3` or `x` terms into two quadratics in this way regardless of whether the resulting quadratics have Real roots.

For example:

`x^4-x^2+1 = (x^2-sqrt(3)x+1)(x^2+sqrt(3)x+1)`