How do you factor rs - rt - ks - ktrsrtkskt?

1 Answer
Apr 22, 2016

rs-rt-ks-ktrsrtkskt 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)rsrtks+kt=(rk)(st)

rs-rt+ks-kt = (r+k)(s-t)rsrt+kskt=(r+k)(st)

but

rs-rt-ks-ktrsrtkskt

cannot be factored further.

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Sketch of a proof

Since all of the terms are of degree 22, then if there is any factorisation, it will be in the form of a product of two factors of degree 11.

Since there are no terms in r^2r2, s^2s2, k^2k2 or t^2t2, the variables rr, ss, tt and kk can each only occur in one factor.

Since there are no terms in rkrk or stst, the pairs r, kr,k and s, ts,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+abktrsrtkskt=(r+ak)(s+bt)=rs+brt+aks+abkt

for some constants aa and bb.

Equating coefficients we find:

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

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

So there is no such factorisation.

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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)