
Polynomial, direct sum
Let P be the vector space consisting of all polynomials with degree less than or equal to 2, where the field that P is over is the set of rationals. Now let E2 be the subspace of all stuff in P that has a root of 2, let E5 be the subspace of all elements in P that have a root of 5. Show that P=E2+E5, where + is the subspace sum.
This isn't a homework problem, but it's a problem that I've been having difficulty with, so any help would be much appreciated. Thanks

EDIT: sorry, i misread what you meant by subspace sum.
So just to clarify the definitions, $\displaystyle P = \{ f(x) \in \mathbb{Q}[x] deg(f(x))\leq 2 \} $
$\displaystyle E2 = \{f(x) \in P f(2)=0 \} $
$\displaystyle E5 = \{f(x) \in P f(5)=0 \} $
$\displaystyle E2 + E5 = \{ f(x)+g(x)  f(x) \in E2, g(x) \in E5 \} $


Any member of E2 is of the form $\displaystyle a(x 2)(x b)= ax^2 a(2+ b)x+ 2ab$ for rational numbers a and b. Any member of E5 is of the form $\displaystyle c(x 5)(x d)= cx^2 c(5+d)x+ 5cd$ for rational numbers c and d. Any polynomial in the direct sum of E2 and E5 is of the form $\displaystyle m(ax^2 a(2+ b)x+ 2ab)+ n(cx^2 c(5+ d)x+ 5cd)= (am+ nc)x^2 (2am+ abm+ 5cn+ cdn)x+ (2ab+ 5cd)$ for rational numbers a, b, c, d, m, and n. Certainly each of the coefficients is a rational number so E2+ E5 is a subspace of P.
Any member of P is of the form $\displaystyle ux^2+ vx+ w$ with u, v, and w rational numbers. If you can show that, given any rational numbers u, v, and w, there exist rational numbers a, b, c, d, m, and n such that am+nc= u, 2am+ abm+ 5cn+ cdn= v, and 2ab+ 5cd= w, you are done. Since that is 3 equations in 6 numbers, it should not be two difficult to show that there exist at least one solution for every u, v, w. (You should NOT expect that solution to be unique.)