Completing the square

Nice Problem Deepansh! $(x-1)(x+3)(x-4)(x-8)=({ x }^{ 2 }-5x+4)({ x }^{ 2 }-5x-24)$ .

Then Let $({ x }^{ 2 }-5x-24)=y$ .

The rest polynomial becomes $y+28$ .

Now we can write this as

$y(y+28)= y^{2}+28y$ .

Now we know to be a square the factors must be equal or we should be able to factorise the above as $(x+a)(x+a)$ .

The above condition is fulfilled only when The last term is 196.

Then we can factorise it as $({ y }^{ 2 }+28y+196)=(y+14)(y+14)$ . Hence we get our solution!

We form two pairs out of the given terms which are multiplied:

$(x-1)(x-4) \text {and} (x-8)(x+3)$ Notice that, on expanding the expression becomes

$(x^2 - 5x +4)(x^2 -5x-24)+m$

Let $x^2 -5x = t$

$\Rightarrow (t+4)(t-24)+m = k^2$

$t^2 -20t -96+m = k^2$

Notice that $t^2-20t+100 = (t-10)^2$

$\Rightarrow x =196$

Note: This solution is specific to this problem only.

$\color{#3D99F6}{(x-1)(x+3)(x-4)(x-8)}+m \quad \quad \small \color{#3D99F6}{\text{May use Vieta's formulas to expand.}} \\ = \color{#3D99F6}{x^4-10x^3+5x^2+100x-96}+m \quad \quad \small \color{#3D99F6}{\text{Divided by }x^2.} \\ = x^2 - 10x +5 + \dfrac{100}x + \dfrac{m-96}{x^2} \\ = \left(x^2 \color{#3D99F6}{- 2\sqrt{m-96}} + \dfrac{m-96}{x^2} \right) - 10 \left( x - \dfrac{10}x\right) + 5 \color{#3D99F6}{+ 2\sqrt{m-96}} \\ = \left(x - \dfrac{\sqrt{\color{#3D99F6}{m-96}}}x \right)^2 - 10 \left( x - \dfrac{\color{#3D99F6}{10}}x\right) + 5 + 2\sqrt{\color{#3D99F6}{m-96}} \quad \quad \small \color{#3D99F6}{\text{Equating }\sqrt{m-96} = 10 \implies m = \boxed{196}} \\ = \left(x - \dfrac{10}x \right)^2 - 2(5) \left( x - \dfrac{10}x\right) + 25 \\ = \color{#3D99F6}{\left(x - \dfrac{10}x - 5 \right)^2 \quad \quad \small \text{A perfect square.}}$

$\implies m = \boxed{196}$ .

@Chew-Seong Cheong sir please upload a solution ...

I don't know if it is a good way to solve, but the constant terms multiply to $-96$ and from adding the options only $196$ gives a perfect square with a $+$ sign.