To the unknown base

Let's evaluate the following wigglish expression :

`\qquad\qquad\qquad 3_{x}4_{x}2_{x}5.`

Telescoping from the left, we get

`\qquad\qquad\qquad 3_{x}4_{x}2_{x}5  =  (3 x  +  4)_{x}2_{x}5,`
`\qquad\qquad\qquad\qquad  =  (3 x^2  +  4 x  +  2)_{x}5,`
`\qquad\qquad\qquad\qquad  =  3 x^3  +  4 x^2  +  2 x  +  5.`

So in brief, we get

`\qquad\qquad\qquad 3_{x}4_{x}2_{x}5  =  3 x^3  +  4 x^2  +  2 x  +  5.`

The pattern that suggests itself here is true in general---a polynomial in the variable `x` can be thought of as a number in base `x`.

Polynomials can be added, subtracted, and multiplied as if they were numbers in base `x`.  For instance

`\qquad\qquad\qquad (3 x^3  +  4 x^2  +  x  +  5)  +  (x^3  +  2 x^2  +  7 x  + 2)`
`\qquad\qquad\qquad\qquad =  4 x^3  +  6 x^2  +  8 x  +  7`

becomes

`\qquad\qquad\qquad 3_{x}4_{x}1_{x}5  +  1_{x}2_{x}7_{x}2`
`\qquad\qquad\qquad\qquad =  4_{x}6_{x}8_{x}7.`

The last calculation resembles the ordinary addition `3415 + 1272  =  4687.`

The product

`\qquad\qquad\qquad (x + 2) (x + 3)  =  x^2 + 5 x + 6`

can be rendered

`\qquad\qquad\qquad (1_{x}2) (1_{x}3)  =  1_{x}5_{x}6

and this resembles the ordinary numerical product `12 xx 13  =  156.`

Base `x` calculations do not always agree with corresponding decimal calculations.  The base `x` calculation is most often actually simpler, because there is no carrying, as there with ordinary numbers, e.g. :

`\qquad\qquad\qquad (15) (15)  =  1 ^{1}0^{2}5  =  225, `

but, (no carrying steps this time),

`\qquad\qquad\qquad (1_{x}5) (1_{x}5)  =  1_{x}10_{x}25.`