$a,b,c \in\mathbb{Z}$ and $x\in\mathbb{R}$, then the following expression is always true:
$$(x-a)(x-6)+3=(x+b)(x+c)$$
Find the sum of all possible values of $b$.
A) $-8$
B) $-12$
C) $-14$
D) $-24$
E) $-16$
I didn't understand what is the meaning of "...is always true".
Even though I can't understand the question, I wrote these:
$$(x-a)(x-6)+3=(x+b)(x+c) \Rightarrow x=\frac{6a-bc+3}{6+a+b+c}$$
Here, $b$ can take an infinite number of values. Or do I miss something? For example, let random values $a=100,b=50,c=3$ then $x=\frac {151}{53}$.
Is there a problem with the question?
To answer the explicit question, "Is there a problem with the question?," the answer is Yes, it's worded in a weird, nonsensical way. (I think this is why Dr. Sonnhard Graubner left a comment asking for the question's source: was it reproduced verbatim, or did the OP paraphrase the problem?) A better version would be something like this: