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

1 Answer
George C.
Apr 22, 2016

#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)#

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