Solve $x^3 +3x -4 =0$ in Real numbers Solve $x^3 +3x -4 =0$ in Real Numbers.
I know the answer is 1 but i want the solution. if you solve it with decomposition, explain how did you find what to do to decompose this expression. (I know the Decomposition but first I find 1 with testing some numbers for x and then by dividing the expression, I find the decomposition but I want a solution that doesn't need testing numbers).
 A: As mentioned in the comments, it is easy to see that $x=1$ is a root.
The derivative of $x^3 +3x -4$ is $3x^2 +3$ which is never zero for $x$ real.
Therefore $f(x)=x^3 +3x -4$ is strictly increasing and so $x=1$ is the only real root.
A: note that $$x^3+3x-4=(x-1)(x^2+x+4)$$
A: For this, I would just factor out an $x-1$. So $x^3+3x-4=(x-1)(x^2+x+4)$. Thus trivially $x=1$ is a real root, and via the quadratic formula (or your method of choice) on the term $x^2+x+4$ you find two other complex roots, but that's not what you wanted so just take the real root.
A: Hint:-

Use factor theorem which states that $(x-n)$ is a factor of $f(x)$ if $f(n)=0$.

Putting $f(1)$ in your polynomial $f(x)=x^3+3x-4$ gives $0$.
Hence $(x-1)$ is a factor.
A: Let $x=u+v$, then $(u+v)^3+3(u+v)-4=0$
$u^3+v^3+3(u+v)(uv+1)=4$
We have one degree of freedom in choosing $u$ or $v$.
Take $uv+1=0$
Now $\displaystyle v=-\frac{1}{u}$ and 
$\displaystyle u^3+v^3=4 \implies u^3-\frac{1}{u^{3}}=4$
$u^6-4u^3-1=0 \implies u^{3}=2\pm \sqrt{5}$
$\displaystyle (u,v)=
\left( \frac{1\pm \sqrt{5}}{2} , \frac{1\mp \sqrt{5}}{2} \right) \implies x=1$
A: Use Cardan's Method to solve cubic  
https://www.math.ucdavis.edu/~kkreith/tutorials/sample.lesson/cardano.html
Also since you know 1 is real root of equation, you can use synthetic division to find other one's
A: 
Hint: It is sometimes helpful to give the integer root theorem a try. Since the polynomial 
  \begin{align*}
p(x)=x^3+3x-4
\end{align*}
  has integral coefficients with $1$ beeing the coefficient of the highest degree, we conclude that if there are integer roots they will divide $-4$, the coefficient of $x^0$.

So, possible candidates are $\pm1,\pm2,\pm4$ from which we obtain already by the first try
\begin{array}{cccc}
p(1)=\color{blue}{0}
\end{array}

Since $1$ is an integer root with $p(1)=0$, we can continue with factorization and obtain
  \begin{align*}
p(x)&=x^3+3x-4\\
&=(x-1)(x^2+x+4)
\end{align*}
  The other roots are those of the quadratic polynomial and can be found using standard techniques.

According to the hint from @user1952009 we see that the discriminant $\Delta$ of the quadratic polynomial $x^2+x+4$ is $$\Delta=b^2-4ac=1-16<0$$ So,  both solutions are not real numbers, leaving $1$ the only real solution.
