Find all 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 and `alpha,beta in I.`

Explanation

We have `(x-a)(x-10)=1=(x+alpha)(x+beta)`
On putting `x=-alpha` in both sides we get
`(-alpha-a)(-alpha-10)+1=0`
`:.(alpha+a)(alpha+10)=-1`
`alpha+a` and `alpha+10` are integers. `[:' a, alpha epsilon]`
`:.alpha+a=-1` and `alpha+10=-1`
i. If `alpha+10=1`
`:.alpha=-9` then `a=8`
Similarly `beta=-9`
Here `(x-8)(x-10)+1=(x-9)^(2)`
(ii) If `alpha+10=-1`
`:. alpha=-11` then `a=12`
Similarly `beta=12`
Here `(x-12)(x-10)+1=(x-11)^(2)`
Hence `a=8,12`