# Help checking whether 3 polynomials generate $P_2(\mathbb{R})$ and $P_3(\mathbb{R})$

Do $x^3 − 2x^2 + 1$, $4x^2−x+ 3$ and $3x − 2$ generate the vector space $P_2(ℝ)$; and $P_3(ℝ)$?

My answer was yes, they do generate $P_3(\mathbb{R})$ and thus also $P_2(\mathbb{R})$. I wrote the coefficients of the polynomials as vectors and then added them together using other coefficients, so I could see what the span was. This gave me:

$S=\{(1, 2, 0, 1), (0, 3, -1, 3), (0, 0, 3, -2)\}$ the coefficients of the polynomials written as vectors.

$\text{span}(S)=\{(a, 2a+4b, -b+3c, a+3b-2c):a,b,c\in\mathbb{R}\}$

Let $a=p$, $2a+4b=q$, $-b+3c=r$, $a+3b-2c=s$.

As $p, q, r, s$ are not multiples of each other, the $4$ elements in $\mathbb{R}$ are variable in all 4 dimensions of $S$, so that means the polynomials generate $P_{4-1}(\mathbb{R}) = P_3(\mathbb{R})$. Since $P_2(\mathbb{R})$ ϵ $P_3(\mathbb{R})$, the polynomials also generate $P_2(\mathbb{R})$.

You are not correct. Note that you can't generate the constant polynomial $1$. In fact the three given polynomials $$x^3 − 2x^2 + 1\;,\;4x^2−x+ 3\;,\;3x − 2$$ have all different degrees greater than zero and therefore any non-zero linear combination has degree greater than $0$.