What Will You Do Now?

Let's analyze the equation $(x+2)(x+3)(x-4)(x-5) = 0$

The sum of its roots is the same as that of the equation of the problem, because subtracting $44$ from the product doesn't change the coefficient of $x^3$ , which is equal to $2+3-4-5 = -4$ , giving $a+b+c+d = 4$

And the product of the roots of the equation above is $2 \times 3 \times 4 \times 5 = 120$ (I ignored the negative signs because there is an even amount of them). But $abcd = 120-44 = 76$

So $a+b+c+d+abcd = 4+76 = \boxed{80}$

Solve the equation and you get $x^{4} - 4x^{3} + 19x^{2} + 46x + 76$ .

Now by Vieta's Theorem Sum of Roots = 4 and Product of roots = 76.

Therefore $\boxed{80}$

(x+2)(x+3)(x-4)(x-5)=44 when expanded, can be written as x^4-4x^3-19x^2+46x+76=0 sum of roots = -(-4)/1=4 product of roots = 76/1 (a+b+c+d)+abcd=4+76=80