An application of Weierstrass theorem? I'm going through some problems and I'm really stumped on this one. The questions says that   

Given $f(x)=|x|$, show that there is a sequence of (real) polynomials $P_n(x)$ with $P_n(0)=0$ that converge uniformly to $f(x)$ on the interval $[-1,1]$. 

I think an application of the Weierstrass theorem is in order, but I don't know how to apply it here and so I'll need some help.
Thanks.
 A: As you correctly guessed, you can use that by Weierstrass's theorem there is a sequence of polynomials $Q_n(x)$ uniformly converging to $f$ on $[-1,+1]$.
Done? Not quite: those $Q_n$'s might not satisfy $Q_n(0)=0$.
Well, then give them a little push that will force them to comply: do you see what you have to add to each of them to obtain $P_n$ and why the push is little (and becomes littler and littler) ?
And do you see why the new sequence of $P_n$'s will still converge to $f$ uniformly  because of the mentioned littleness?
And do you see that the exact formula for  $f$ is a red herring and that only the fact that $f(0)=0$ is relevant?     
Yes, I'm sure you'll see all that after a short moment a reflexion. Good luck!
A: As some proofs of the Weierstrass Approximation Theorem use this result for its proof, I am somewhat unsatisifed with using the Weierstrass Approximation Theorem to prove it.
The Taylor series for $\sqrt{1-x}$ is
$$
\sqrt{1-x }=1-{1\over 2}x-{1\over 2\cdot4}x^2-{1\cdot3\over 2\cdot4\cdot6}x^3-\cdots.
$$
This converges uniformly  for $0\le x\le 1$.
From this, we may represent $|x|$ with (replace "$x$" with "$1-x^2$")
$$
|x|=1-{1\over 2}(1-x^2)-{1\over 2\cdot4}(1-x^2)^2-{1\cdot3\over 2\cdot4\cdot6}(1-x^2)^3-\cdots.
$$
This will converge uniformly for $0\le1-x^2\le  1$; that is, for $-1\le x\le 1$.
You can then modify things to produce polynomials $P_n$ with $P_n(0)=0$ that converge uniformly to $|x| $ on $[-1,1]$.
A: Actually you can use Bernstein polynomial formula to approximate any continous functions on [a,b].
