Mmm Tasty

$(x-1)(x-4)(x+3)(x-8)+m\\=(x^2-5x+4)(x^2-5x-24)+m$ Let $y=x^2-5x+4$ : $y(y-28)+m\\=y^2-28y+m$ In order to fit the form of the square of a binomial $(a+b)^2=a^2+2ab+b^2$ , $m$ must be equal to $\boxed{196}$

Very interesting problem, I got this problem wrong the first try because I accidentally made my last coefficient positive so my original answer was 4.

Anyway, because the problem has to be a perfect square, it can be factored into:

$(x^2+bx+c)^2$

or

$a^2x^4+2 a b x^3+b^2 x^2+2 b c x+2 a c x^2+c^2$ .

After some expansion of our original equation, we get $x^4-10x^3+5x^2+100x-96$ thus $\text{a= 1,b= -5, c= -10}$ . Now, because m is a constant, we only need to focus on the last coefficient. Finally solving for m:

$m=c^2-(-1)(3)(-4)(-8)\Rightarrow 100+96=\boxed{196}$ .