Show that $2x^3+2x^2-10x+6$ is positive if $x>1$ 
Show that $2x^3+2x^2-10x+6$ is positive if $x>1$ 

I have tried to solve it using mean value theorem but it doesn't work. Please anyone help me to solve this.
 A: Alt. hint:  let $\,x-1 = y \gt 0\,$, then:
$$
2x^3+2x^2-10x+6 = 2(y+1)^3+2(y+1)^2-10(y+1)+6=2 y^3 + 8y^2 \gt 0
$$
A: Hint. Let $f(x)=2x^3+2x^2-10x+6$. Trying to use the Mean Value Theorem, you computed the derivative $f'$. Note that $f(1)=f'(1)=0$ which implies that $1$ is a double root of $f$, that is the polynomial $f$ is divisible by $(x-1)^2$. Now you should be able to get the complete factorization of $f$ and to show the given inequality.
A: $f'(x)=6x^2+4x-10 >0$ for 
$x>1$, implies
$f$ stricly increasing for $x >1$.
Since $f(1)=0$, we have $f(x) >0$ for $x>1$.
A: Let $$f(x)=2x^3+2x^2-10x+6$$ then $$f'(x)=6x²+4x-10$$ $$=(6x+10)(x-1)$$Now $f(x)$ is strictly increasing if $f'(x)>0$ or when $x$ belongs to $(-\infty , -\dfrac{10}{6}) \cup (1,\infty)$.
At $x=1$, $f(x)=0$ and $f(x)$ is strictly increasing in the interval $(1,\infty)$ so for values of $x$ greater than $1$, $f(x)$ will be greater than $0$ i.e. positive.
A: 
As you can see, given function has a root at x=1. For x>1, function is increasing.
If you take derivative of function, it becomes f'(x)=6$x^2$+4x-10, which is positive for x>1.
 Therefore, function f(x) is increasing, (and positive since f(1)=0) for x>1. 
