# A vector space of the polynomials with given root

Is the set of the real polynomials, which have as root number 2, a vector space?

My try:

Well, a polynomial from that given set could look like:

$p(x) = (x-2)(a_0x^0 + a_1x^1 + a_2x^2+...+a_nx^n)$

So if I have two polynomials from that given set, then $p_1(x) + p_2(x)$ must be a polynomial with that root and it has to have the same degree, in our case that degree is $n$.

So, $p_1(x) + p_2(x) = (x-2)(a_0x^0 + a_1x^1 + a_2x^2+...+a_nx^n) + (x-2)(b_0x^0 + b_1x^1 + b_2x^2+...+b_nx^n) = (x-2)[(a_0+b_0)x^0 + (a_1+b_1)x^1 + (a_2+b_2)x^2 ... (a_n+b_n)x^n]$.

And if I have a scalar, $r$ from the real numbers, that polynomials need to satisfy:

$rp(x) = r (x-2)(a_0x^0 + a_1x^1 + a_2x^2+...+a_nx^n) = (x-2)(ra_0x^0 + ra_1x^1 + ra_2x^2+...+ra_nx^n)$, so that is ok too.

• Why does the degree have to be the same? Commented Feb 6, 2017 at 19:47
• It doesn't have to?
– Leif
Commented Feb 6, 2017 at 19:51
• What about x-2? Commented Feb 6, 2017 at 20:38
• What about zero? Commented Feb 6, 2017 at 20:39

Why degrees of $p_1,p_2$ should be the same?
Let $p_i(x)=(x-2)q_i(x)$ and $\alpha,\beta\in\Bbb R$. Then $\alpha p_1(x)+\beta p_2(x)=(x-2)\bigl(\alpha q_1(x)+\beta q_2(x)\bigr)$. Thus the set under consideration is a vector subspace of the vector space of all polynomials, so it is itself the vector space.