The number of integral values of $a$ for which the quadratic expression $(x-a)(x-10)+1$ can be factored as a product $(x+\alpha)(x+\beta)$ of two factors $\alpha, \beta, \in I$, is

Expanding the expression gives x squared minus (a plus 10)x plus (10a plus 1). Since this factors into (x plus alpha)(x plus beta), we have alpha plus beta equals negative (a plus 10) and alpha times beta equals 10a plus 1. Substituting a equals negative 10 minus alpha minus beta into the product equation gives alpha times beta equals negative 10 times (10 plus alpha plus beta) plus 1, which rearranges to alpha times beta minus 10 alpha minus 10 beta equals negative 99. Adding 100 to both sides allows factoring as (alpha minus 10)(beta minus 10) equals 1. Because alpha and beta are integers, the only factor pairs of 1 are (1, 1) and (negative 1, negative 1), yielding exactly two valid values for a. The result is 2.

• Thanks
• Wrong explanation