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$a,b,c \in\mathbb{Z}$ 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?

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    $\begingroup$ I think there is a problem with the question. Do you have the source of it? $\endgroup$ – Dr. Sonnhard Graubner Jan 6 at 12:00
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    $\begingroup$ Hint: Equate the coefficients of $x$. The "are always true" means that the equality holds for every $x$ you choose. The problem with your calculation for $x$ is that the denominator of that fraction vanishes. $\endgroup$ – Michael Burr Jan 6 at 12:03
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    $\begingroup$ Here are a few details: $$x^2-(a+6)x+6a+3=x^2+(b+c)x+bc$$ and for this to be true for ALL values of $x$ we must have equal coefficients on both sides. $\endgroup$ – String Jan 6 at 12:09
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    $\begingroup$ @String $6+a+b+c=0$, and $6a+3-bc=0$ ..?? $\endgroup$ – Elementary Jan 6 at 12:13
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    $\begingroup$ @Beginner: Yes, and then you must find integer solutions and count values of $b$. $\endgroup$ – String Jan 6 at 12:14
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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:

Consider the set of triples $(a,b,c)\in\mathbb{Z^3}$ for which the equation

$$(x-a)(x-6)+3=(x+b)(x+c)$$

holds for all $x\in\mathbb{R}$. Find the sum of all the $b$'s among these triples.

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    $\begingroup$ Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all. $\endgroup$ – David Z Jan 7 at 3:45
  • $\begingroup$ @DavidZ, excellent point! $\endgroup$ – Barry Cipra Jan 7 at 13:02
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The question is poorly worded. It should read something like this:

$a,b,c$ are integers such that the following equation holds for all $x\in\Bbb R$:

etc.

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Two polynomials which are always equal over the reals are exactly the same. In this case, since $x$ is allowed to vary, while $a,b,c$ are fixed, these are two polynomials in $x$.

For them to be equal, the coefficients of $x$ must also be equal. Therefore, \begin{align} -a-6&=b+c\\ 6a+3&=bc. \end{align} Now, you can solve for $a$ in the first equation and substitute into the second equation, giving $$ 6(-b-c-6)+3=bc. $$ The problem then becomes, for which integers does this equation have a solution?

If you solve for $b$ here, you'll get a fraction in $c$, which you can study to figure out which integers for $c$ result in integers for $b$.

The problem with your solution for $x$ is that the denominator of your fraction is zero.

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  • $\begingroup$ You forgot a "$+ 3$" on the left side of your final equation. $\endgroup$ – John Omielan Jan 6 at 12:16
  • $\begingroup$ @JohnOmielan Thanks, that's what I get for answering on my phone. $\endgroup$ – Michael Burr Jan 6 at 12:18
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    $\begingroup$ @Beginner: I would have put it as "Given some $a,b,c\in \mathbb Z$ the following equality holds for all $x\in\mathbb R$ ..." $\endgroup$ – String Jan 6 at 12:22
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    $\begingroup$ I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression. $\endgroup$ – Michael Burr Jan 6 at 12:47
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    $\begingroup$ The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers. $\endgroup$ – ypercubeᵀᴹ Jan 7 at 8:17

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