Let $x$ be a real number. If $x^3 + 7x^2 − 8 \leq 0$, then $x \leq 1$. I was just wondering if this is a good way to write out this proof statement.  I used proof by contradiction.  I am pretty new at using this so I was just wondering if anyone on here could give me advice on the formatting of my proof.  
PROOF: Suppose the opposite, $x^3+7x^2-8 \leq 0$ and $x>1$.
If $P$ and $\neg Q$ for all $x>1$, then it should be true for $x=2$.
Thus, $(2)^3 + 7(2)^2 - 8 \leq 0 \implies 8 + 28 - 8  \leq 0 \implies 28$ is not $\leq 0$.
Therefore, $P$ and $\neq Q$ is false. $\square$ 
 A: Note, as a part of the definition of the $<$ sign and positive/negative numbers one has the following property:

Given $a>0$ and $b>c$ it follows that $ab>ac$

In particular then, supposing $x>1$ one has
$x^2=x\cdot x> x\cdot 1=x$
Applying this once again, we have $x^3>x$ for all $x>1$

Suppose for the sake of proof by contrapositive that $x>1$
It follows then that $x^3>x>1$ and $7x^2>7x>7$ and therefore we have
$x^3+7x-8>1+7-8=0$
This proves then by contrapositive that $x^3+7x-8\leq 0$ implies $x\leq 1$
A: Before guessing the correct language, you need to understand how the question relates to the graph. Since you can factor out the $x-1$ and get a quadratic, you can use the quadratic formula to find the other two roots. 
When the function (the $y$ value) is negative or $0,$ we do have $x \leq 1.$  On the other hand, out of that, $y$ is positive for $x$ between the two roots with negative $x.$ 

A: " If P and ¬Q for all x>1, then it should be true for x=2"
That will be a single confirmation.  That is not enough to prove anything. To prove something you have to show it is true for ALL $x > 1$.  Not just for one value of $x > 1$.   
Suppose I wanted to prove $n^2$ is an even number for all $n \ge 7$.  So I try it with $n=8$ and $n^2 = 8^2 = 64$ is even.  That's not enough to prove it....  Or if I wanted to prove that everyone in New York is named Norman... and I find that one person in New York is named Norman....
But we can prove it for every $x > 1$.  If $x > 1$ then $x^2 = x*x > x*1 =x$ and $x^3 = x*x^2 > x*1 = x > 1$.  And $x > 1 \implies 7x > 7*1=7$.  So $x^3 + 7x -8 > 1 + 7 - 8 > 0$. 
You were correct about the contrapositive.
$x > 1 \implies x^3 + 7x - 8 > 0$ so $x^3 + 7x -8 \le 0 \implies x \le 1$.
