Question

How do you factor `rs - rt - ks - kt`?

Answer 1

`rs-rt-ks-kt` cannot be factored into linear factors.

Explanation:

This is an interesting question in that it looks like a trick question or a typo.

For example,

`rs-rt-ks+kt = (r-k)(s-t)`

`rs-rt+ks-kt = (r+k)(s-t)`

but

`rs-rt-ks-kt`

cannot be factored further.

`color(white)()`
Sketch of a proof

Since all of the terms are of degree `2`, then if there is any factorisation, it will be in the form of a product of two factors of degree `1`.

Since there are no terms in `r^2`, `s^2`, `k^2` or `t^2`, the variables `r`, `s`, `t` and `k` can each only occur in one factor.

Since there are no terms in `rk` or `st`, the pairs `r, k` and `s, t` must each occur in the same factor.

Hence up to scalar factors, the factorisation must be expressible in the form:

`rs-rt-ks-kt = (r+ak)(s+bt) = rs+brt+aks+abkt`

for some constants `a` and `b`.

Equating coefficients we find:

`{ (b = -1), (a = -1), (ab = -1) :}`

which is inconsistent, since `-1 * -1 = 1 != -1`

So there is no such factorisation.

`color(white)()`
Random Advanced Footnote

It is actually possible to factor `rs-rt-ks-kt`, but only over a field of characteristic `2`, not over ordinary numbers. In a field of characteristic `2`, addition and subtraction are the same thing, so we could write:

`rs-rt-ks-kt = rs+rt+ks+kt = (r+k)(s+t)`