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

Apr 22, 2016

$r s - r t - k s - k t$ 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,

$r s - r t - k s + k t = \left(r - k\right) \left(s - t\right)$

$r s - r t + k s - k t = \left(r + k\right) \left(s - t\right)$

but

$r s - r t - k s - k t$

cannot be factored further.

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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 $r k$ or $s t$, 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:

$r s - r t - k s - k t = \left(r + a k\right) \left(s + b t\right) = r s + b r t + a k s + a b k t$

for some constants $a$ and $b$.

Equating coefficients we find:

$\left\{\begin{matrix}b = - 1 \\ a = - 1 \\ a b = - 1\end{matrix}\right.$

which is inconsistent, since $- 1 \cdot - 1 = 1 \ne - 1$

So there is no such factorisation.

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It is actually possible to factor $r s - r t - k s - k t$, 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:
$r s - r t - k s - k t = r s + r t + k s + k t = \left(r + k\right) \left(s + t\right)$