If $(x+1)^2 - 2x>2(x+1) +2$, then $x$ cannot equal which of the following?

Question

See: Nova GRE Math Bible. Page-$235$.

Problem #$10$.
If $(x+1)^2 - 2x>2(x+1) +2$, then $x$ cannot equal which of the following?
(A) $-5$
(B) $-3$
(C) $0$
(D) $3$
(E) $5$

$(x+1)^2 - 2x>2(x+1)+2$
$\Rightarrow x^2+2x+1-2x>2x+4$
$\Rightarrow x^2>2x+4-1$
$\Rightarrow x^2>2x+3$
$\Rightarrow x^2-2x-3>0$
$\Rightarrow x^2-3x+x-3>0$
$\Rightarrow x(x-3)+1(x-3)>0$
$\Rightarrow (x+1)(x-3)>0$

So, $x>-1$, and $x>3$.

Since $-5$ and $-3$ are outside the range of number-line, the answer is:

• (A) and (B).

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But, the correct answer is, (C), and (D).

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What is wrong with my solution?

Answer 1

$$(x+1)^2-2x>2(x+1)+2\iff (x+1)^2>4(x+1)\iff \begin{cases} x+1>4&if\ x>-1\\ x+1<4& if \ x<-1 \end{cases}$$ $$\iff x>3\text{ or }x<-1,$$

Only (C) and (D) are correct.

Answer 2

Everything is fine until you conclude from $(x+1)(x−3)>0$ that $x>-1$ and $x>3$.

Rather, if $(x+1)(x-3)>0$, then $x+1$ and $x-3$ are both positive or both negative.

Answer 3

$(x+1)(x-3)>0$

There are two possible cases:

Case 1: Both $x+1>0$ and $x-3>0$ are true. This happens if and only if $x>3$.

Case 2: Both $x+1<0$ and $x-3<0$ are true. This happens if and only if $x<-1$.

The answer should be $x<-1$ or $x>3$.