Answer:
The set is a basis for P2
Explanation:
Remember, a set B is a basis for a vector space V if B is a linear independent set and gen(B)=V. Also, remember that the polynomials
are linearly independent if the unique scalars
such that

are

Expanding the above equation we have that

Then we need values of a, b and c such that

So, we have a homogeneous system with augmented matrix
![A= \left[\begin{array}{ccc}a&4b&5c\\2a&b&4c\\-3a&-b&-4c\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/3ol111hs61m69t7ryhzcu82rioq76t94cy.png)
Now, we apply row operations for solve the system
1. To the second row of A we subtract the first row twice. And to the third row we add the first row three times. We get the matrix:
![\left[\begin{array}{ccc}a&4b&5c\\0&-7b&-6c\\0&11b&11c\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/i9g6h281dn4e5b117vvgmlbc3fuo0xtnnm.png)
2. We multiply the third row of the previous matrix by 1/11
![\left[\begin{array}{ccc}a&4b&5c\\0&-7b&-6c\\0&b&c\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/b4begsui988rk5qrxji7ii8qycpzlnbhje.png)
3. To the third row of the previous matrix we add 7 times the second row.
![\left[\begin{array}{ccc}a&4b&5c\\0&-7b&-6c\\0&0&42c\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/mi9ajv4sz2ioru9siozu6jwxxjuavq3rsm.png)
Since the rank of the matrix is 3 then the system has unique solution. But the system is homogeneous, therefore the solution is

This show that the polynomials are linear independent and therefore the set is a basis for P2