How can I find basis of polynomial? $$
V=\operatorname{span}\{v(1), v(2), v(3)\}\qquad
W=\operatorname{span}\{w(1), w(2), w(3)\}
$$
where
$$\eqalign{v(1)&=x^3 + 4x^2 - x +3
\\
v(2)&=x^3 + 5x^2     +5
\\
v(3)&=3x^3 + 10x^2 -5x + 5 
\\
w(1)&=2x^3 + 3x^2 -3x + 9
\\
w(2)&=x^3 + 2x^2 -x + 5
\\
w(3)&=x^3 + 4x^2   +6}$$
Find basis for $V, W, V+W, V\cap W$
I founded that $v(3)$ is a linear combination of $v(1)$ and $v(2)$. and $v(1)$, $v(2)$ are linearly independent. Also, $w(1)$, $w(2)$, $w(3)$ are linearly independent. But I cannot find basis for $V, W$. I know that $V+W$ is a linear combination of basis of $V$ and $W$.
When a vector in $R(n)$ is given, I can find basis of $V, W, V+W, V\cap W.$
But I cannot find basis of polynomial. Would you help find basis of polynomial?
 A: It's a well-known result that every linearly independent set can be extended to a basis (via the inclusion of vectors) and that every spanning set can be reduced to a basis (via the omission of vectors). Therefore, if you have anything extra given in your definition of $V$, you can simply get rid of it.  With $W$, you're already done.
$V + W$ I would describe as every linear combination of vectors in $V$ and $W$. Start with your bases from above. Since $P_3(x)$ is a four-dimensional vector space, you'll find that you have one or two extra (i.e., linearly dependent) vectors. We know this because
$$\dim(V+W) = \dim V + \dim W - \dim(V\cap W)$$
which, by the way, should get you started on the last part. Let me know if you have questions.
Edit: Regardless of the questions you may or may not have, I've woken up a bit more, and I think this seems a bit incomplete...
However you determined the independence of vectors for $V$ and $W$, do the same for $V+W$. You'll find that $V+W = P_3(x)$. Therefore, $\dim (V+W) = 1$. So we need to find a polynomial that's in both $V$ and $W$.
My TI-83 tells me that I can write $V= \text{span}\left(x^3-5x-5, x^2+x+2\right)$ and $W = \text{span}\left(x^3+6, x^2, x+1 \right)$. Any polynomial that's in both, then, can be written as a linear combination of either set of vectors. In other words,
$$
\begin{aligned}
\alpha \left( x^3 - 5x - 5\right) + \beta \left( x^2 + x + 2\right) &= \gamma \left( x^3 + 6 \right) + \delta \left( x^2 \right) + \epsilon \left( x + 1 \right) \\
\ \alpha x^3 + \beta x^2 + \left( \beta - 5\alpha \right) x + \left( 2\beta - 5\alpha \right) &= \gamma x^3 + \delta x^2 + \epsilon x + \left( 6\gamma + \epsilon \right) \\
\end{aligned}
$$
Since these two polynomials are equal, their coefficients are equal. And so we have the system of equations
$$
\begin{aligned}
\alpha &= \gamma \\
\beta &= \delta \\
\beta - 5\alpha &= \epsilon \\
2\beta - 5\alpha &= 6\gamma + \epsilon \\
\end{aligned}
$$
From which it follows quickly that $\beta = 6\alpha$. This will suggest that a specific polynomial (or more precisely, any scalar multiple of a specific polynomial) resides in both $V$ and $W$. Which one is it?
