Verified answer

the condition of passing through the origin means that the zero degree term is missing

if we represent a polynomial as a 5-ple of scalars, the coefficients of the polynomial, in descending order

the last component is 0

IMO it is a vector space, indeed:

polynomial addition is an internal operation

(a,b,c,d,0) + (e,f,g,h,0) = (a+e,b+f,c+g,d+h,0)

multiplication by a scalar is internal as well

t(a,b,c,d,0) = (at,bt,ct,dt,0)

associativity and commutativity hold

it has the null vector (0,0,0,0,0)

each vector (a,b,c,d,0) has its inverse (-a,-b,-c,-d,0)

distributivity works in both cases

1)

k((a,b,c,d,0) + (e,f,g,h,0) ) = k(a+e,b+f,c+g,d+h,0) =

= (ka+ke,kb+kf,kc+kg,kd+kh,0) = (ka,kb,kc,kd,0) + (ke,kf,kg,kh,0) =

= k(a,b,c,d,0) + k(e,f,g,h,0)

2)

(k+h)(a,b,c,d,0) = k(a,b,c,d,0) + h(a,b,c,d,0)

and also

h(k(a,b,c,d,0)) = (hk)(a,b,c,d,0)

and unity of scalar 1(a,b,c,d,0) = (a,b,c,d,0)

all axioms work