Let W be the set of all vectors p(t) in P such that p(0)=0. Show...

A polynomial ${\displaystyle p\left(x\right)=a{x}^2+bx+c}$ satisfies ${\displaystyle p\left(1\right)=0}$ if and only if ${\displaystyle a+b+c=0}$
Hence every ${\displaystyle p\in W}$ is of the form
${\displaystyle p\left(x\right)=a{x}^2+bx+(-a-b)=a(x^2-1)+b(x-1)}$ (1)
Can you use equation (1) to find a basis for W and thus compute dimW?
Sidenote: There is a more fun way to do this problem. The map ${\displaystyle T:P_2\Rightarrow R}$ given by ${\displaystyle T\left(p\right)=p\left(1\right)}$ is linear and ${\displaystyle kerT=W}$. Since ${\displaystyle T\left(1\right)=1}$ it is clear that T has rank 1. The Rank-Nullity theorem then gives dimW.

a.Since (1)=0, x-1 is a factor of P. So any P in W is of the form: . So if P and S are two elements in W, then, (P+S)(1)=0+0=0. So $P+S\in W$, and $rP(1)=r\cdot 0=0$. So $rP\in W$ for any real number r. So W is a subspace of $P_2$
b. To get a dimension of W we find a spanning set T of W and independent as well. Consider $T=\{x-1,(x-1{)}^2\}$. T is independent. We need to show T spans W. Let P be any element of W. So $P(x)=(x-1)\cdot q(x)$. If q(x) is a scalar, say $q(x)=m,thenP=m(x-1)+0(x-1{)}^2$, and if q(x) is not a scalar, then $q(x)=ax+b$ for some number a,b. So $P(x)=(x-1)(ax+b)=(b-a)(x-1)+a(x-1{)}^2$. So T spans W. Thus: T is a basis with 2 elements. So $dim(W)=2$