What is the dimension of the space of all polynomials $f(t)$ in $\mathrm{Pol}_2(\mathbb{R})$ such that $f(1) = 0$?
So we have some arbitrary equation $xt^2 + yt + z = 0$. Would the dimension of this space simply be 2=3-1 because it takes away one degree of freedom ($f(1) = 0$)?
On
Let's first ask the question what dimension the vektorspace $P2$ has over your given field $K$. Here I'd assume that $P2=\{f\in K[T]\,|\,\deg(f)\le 2\}$. Within $P2$ you can find the linearly independent elements $T^0,T^1,T^2$, which is also a generator set, seeing that any polynomial from degree $\le 2$ can be written as $a_2 T^2+a_1 T^1+a_0 T^0$ with $a_0,\dots ,a_2\in K$. Now this implies that your set $P2$ has dimension $3$ over the arbitrary field $K$.
Now we gotta think about the second condition forced upon us, which is $f(1)=0$. Let's first think about what polynomials are actually given by this equation: Since $f(1)=0$, we know that the polynomial $T-1$ divides into $f$, i.e. $f(T)=(T-1)g(T)$, where $g$ needs to be a polynomial of degree $<2$. Then again, any polynomial $g$ with degree $\deg(g)\le 1$ satisfies that for $f(T):=g(T)\cdot(T-1)$ the above condition holds, i.e. you can use any polynomial $g\in P1$ to define a polynomial with the above condition.
Now we have found that any of your polynomials can be written as $(T-1)\cdot g(T)$ with $g\in P1$ and also for any $g\in P1$ the polynomial $(T-1)\cdot g(T)$ satisfies your proposed condition. It is only natural to assume that the dimension must be the same as $P1$, which is $2$ with an analogous argument to the first paragraph.
This assumption must be proven first, which requires you to show that for any two linearly independent $g_1,g_2\in P1$ the elements $(T-1)g_1(T)$ and $(T-1)g_2(T)$ are also linearly independent. Can you spot why this would suffice without the need to show that those $2$ linearly independent elements also generate your entire set?
So, I answer your question given some conditions. First, I assume you are talking about a subspace of $\mathrm{Pol}_2(\mathbb{R})$, that is all real polynomial functions of degree less than two.
The space you described is $U=\{f\in\mathrm{Pol}_2(\mathbb{R})\mid f(1)=0\}$, i.e. all polynomials having a root at $1$, i.e. $U=\{f(t)=at^2+bt+c\in\mathrm{Pol}_2(\mathbb{R})\mid f(1)=a+b+c=0\}$.
Note first, that this is really a subspace, i.e. the null polynomial $\mathbf0\in U$, as $\mathbf0(t)=0$ f.a. $t\in\mathbb{R}$. Second, if $f,g\in U$, then $(f+g)(1)=f(1)+g(1)=0$, i.e. $f+g\in U$ for the canonical(pointwise addition in $\mathrm{Pol}_2(\mathbb{R})$). Lastly, for $f\in U$, we have $(\lambda f)(1)=\lambda\cdot f(1)=\lambda\cdot 0=0$.
Now, the canonical basis for $\mathrm{Pol}_2(\mathbb{R})$ stems from its correspondence with $\mathbb{R}^3$ and is given as $S=(1,x,x^2)$. Followingly, the dimension of $\mathrm{Pol}_2(\mathbb{R})$ is $3$. Note that the vectors in this basis are functions. Check maybe yourself that this basis really is linearly independent and spans $\mathrm{Pol}_2(\mathbb{R})$(try to express some $f(t)=at^2+bt+c$ as a linear combination of $S$).
You can now solve the equation $a+b+c=0$, defining $U$ over a condition on the coefficients of a corresponding polynomial $f(t)=at^2+bt+c$.
For this, let $c$ and $b$ just be constants, then we can express $a=-b-c$. This gives us an arbitrary choice of the constant $c$ and $b$, with $a$ resulting from these two. Followingly, the dimension of your space is $2$.
A corresponding basis would be $(-x^2+x,-x^2+1)$. This stems again from the correspondence of $\mathbb{Pol}_2(\mathbb{R})$ and $\mathbb{R}^3$ as the corresponding space there, $\{(x,y,z)\in\mathbb{R}^3\mid x+y+z=0\}$ has a similar basis of $((-1,1,0),(-1,0,1))$.