Writing an equation with integer coefficients An algebraic number is any complex number that is a root of a non-zero polynomial in one variable with rational coefficients. for example write an equation which $\sqrt2 +1$ is  one of its root ,with integer coefficients .
$$x=\sqrt 2+1 \to x-1=\sqrt 2\\(x-1=\sqrt 2)^2\to \\x^2+1-2x=2\\x^2-2x-1=0$$ 
other example :write an eq. with integer coefficients which $x=\sqrt2 +\sqrt3$ is a root.
$$x=\sqrt2 +\sqrt3 \\x^2=2+3+2\sqrt6\\(x^2-5)=(2\sqrt6)\\(x^2-5)^2=(2\sqrt6)^2\\x^4+25-10x^2=24\\x^4-10x^2-1=0$$
or$$x-\sqrt2=\sqrt3\\x^2+2-2\sqrt2x=3\\x^2-1=2\sqrt2x\\(x^2-1)^2=(2\sqrt2x)^2\\x^4+1-2x^2=8x^2\\x^4-6x^2+1=0$$ 
Now: my question is about :an equation which one's root is $x=\sqrt2+\sqrt3+\sqrt 5$ and coefficients are integer . 
How many equation we can write for it ? w.r.t "coefficients are integer"
How can write this equation easily ?
Thanks in advance.

One way is :$$x=\sqrt2+\sqrt3+\sqrt 5\\x-\sqrt5=\sqrt2+\sqrt3\\(x-9\sqrt5=\sqrt2+\sqrt3)^2\\x^2+5-2\sqrt5x=2+3+2\sqrt6\\x^2-2\sqrt5x=2\sqrt6\\(x^2-2\sqrt5x=2\sqrt6)^2\\x^4+20x^2-4\sqrt5x^3=24\\(x^4+20x^2-24)^2=(4\sqrt5x^3)^2$$ Is there an easier way  ?
