Show that polynomials with a given factor form a subspace

Question

I have a question for one of my assignments but I don't understand how to solve it.

Let $P_n$ be the set of real polynomials of degree at most $n$, show that

$S=\{p ∈ P_7:x^2+x+4 $ is a factor of $p(x)\}$ is a subspace of $P_7$

I know that to prove it is a subspace I need to show that it contains the zero vector, is closed under vector addition, and is closed under scalar multiplication. I know that it satisfies all those conditions because they are given, but I'm now sure how to prove it because I'm only given the factor.

A concern I have with the solution posted on a previous post with a similar question was to introduce a function $f$ such that $p=q*f$ where $q$ is the given factor. If $f$ has a non-zero coefficient for the $x^0$ factor, then that would mean it wouldn't include the zero vector

Edit: It is not given that the conditions are satisfied, but I am able to check if they are and I'm using the that function so see if it is. I am supposed to prove that satisfies those conditions without the knowing that it is.

Answer 1

To say that $x^2+x+4$ is a factor of the polynomial $p(x)$ simply means that $p(x)$ is expressible as $$ p(x)=(x^2+x+4)q(x) $$ for some other polynomial $q(x)$. This is exactly akin to saying $5$ is a factor of $40$ because $40 = 5 \times 8$.

So, there is a straightforward "rule of membership" for your set $S$: the polynomial $p(x)$ of degree $7$ or less belongs to $S$ is and only if $p(x)=(x^2+x+4)q(x)$ for some other polynomial $q(x)$. [Don't worry about the degree of $q$ right now.]

Now we can easily check the subspace conditions. First, note that $p(x)=0$ meets the "rule of membership" for $S$: $0 = (x^2+x+4)\cdot 0$ [so just take $q(x)=0$, a valid choice]. Thus $0 \in S$.

Now suppose $f(x)$ and $g(x)$ meet the "rule of membership" for $S$: they are each degree $7$ or less, and $f(x)=(x^2+x+4)q(x)$ and $g(x) = (x^2+x+4)r(x)$ for some mystery polynomials $q(x)$ and $r(x)$. Clearly, $f(x)+g(x)$ still has degree $7$ or less. Moreover, $$ f(x)+g(x) = (x^2+x+4)q(x) + (x^2+x+4)r(x) = (x^2+x+4)(q(x)+r(x)). $$ As $q(x)+r(x)$ is still a polynomial, we've shown that $f(x)+g(x)$ meets the "rule of membership" for $S$. Hence $f(x)+g(x) \in S$ so $S$ is closed under addition.

You can now prove closure under scalar multiplication using the exact same strategy: just check the rule of membership for $cf(x)$ if $f(x) \in S$ and $c$ is a constant.