$a, b, c, d$ are distinct integers such that $(x - a)(x - b)(x - c)(x - d) = 4$ has an integral root $r$. Then $a + b + c + d$ is equal to:
Step-by-Step Solution
Key Concept: The sum of the four distinct integer factors of 4 that appear as $(r-a), (r-b), (r-c), (r-d)$ is zero, so $a+b+c+d = 4r - 0 = 4r$.
Let $r$ be an integral root of $(x-a)(x-b)(x-c)(x-d)=4$. Then $(r-a)(r-b)(r-c)(r-d)=4$. Since $a,b,c,d$ are distinct integers and $r$ is also an integer, each factor $(r-a), (r-b), (r-c), (r-d)$ is a non-zero integer. We need four distinct integers whose product is 4. The only factorizations are $4=1\cdot 1\cdot 1\cdot 4$ (not distinct) or $4=2\cdot 2\cdot 1\cdot 1$ (not distinct) or $4=(-1)\cdot(-1)\cdot(-2)\cdot 2$ (distinct). So $(r-a), (r-b), (r-c), (r-d)$ equal $-2,-1,1,2$ in some order. Therefore $a+b+c+d = 4r-(2+1-1-2) = 4r$.
Correct Answer: 4