Question

Let V be the vector space P3[x] of polynomials in x with degree less than 3 and W be the subspace

Answer 1

By definition, the span of two vectors is the set of all possible linear combinations of said vectors. So, any element in W is obtaining by choosing two numbers a,b and building

$a(7-8x-8x^2)+b(x^2-(5+6x)) = -8ax^2-8ax+7a+bx^2-6bx-5b$

(a)

So, you can show a nonzero polynomial in W by choosing any values of a and b, as long as they're not both zero. To keep it as simple as possible, we can choose for example a=1, b=0 and we obtain one of the base vectors of W:

$7-8x-8x^2$

(b)

Factoring the powers of x, we see that a generic polynomial in W looks like

$(b-8a)x^2-(8a+6b)x+7a-5b$

So, we must build a polynomial

$dx^2+ex+f$

such that it is not possible that

$\begin{cases}d=b-8a\\e=8a+6b\\f=7a-5b\end{cases}$

For example, let's try the polynomial $x^2+x+1$

We should solve the system

$\begin{cases}b-8a=1\\8a+6b=1\\7a-5b=1\end{cases}$

And you can check that it has no solutions. So, the polynomial $x^2+x+1$ does not belong to W. On the other hand, it surely belongs to V, because it is a polynomial with degree less than 3. So, the polynomial $x^2+x+1$ belongs to V/W.