Polynomial prove exercise $P(x)=x^n + a_1x^{n-1} +\dots+a_{n-1}x + 1$ with non-negative coefficients has $n$ real roots. Prove that $P(2)\ge 3n$ I don't have an idea how to do that, I'm in 4th grade high school, you don't have to solve it for me, I'd be just happy to get a clue how - + if someone could provide me with some sort of exercises like that with explanation I'd be happy thanks.

Since $a_j\geq 0$ $(1\leq j\leq n-1)$, all zeros are negative. Hence
$$ P(x)=(x+r_1)\cdots (x+r_n)\qquad r_j>0, 1\leq j\leq n$$
and $r_1\cdots r_n=1$. It follows from this last equality that at least one $r_j$ is greater than or equal to $1$, which gives $r_1+\cdots+r_n\geq 1$.
It follows that
$$ P(2)=(2+r_1)\cdots (2+r_n)\geq 2^n+(r_1+\cdots+r_n)2^{n-1}+1\geq 2^n+2^{n-1}+1\geq 3n.$$
